How to find the surface area of a cylinder - SAT Math
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An upright cylinder with a height of 30 and a radius of 5 is in a big tub being filled with oil. If only the top 10% of the cylinder is visible, what is the surface area of the submerged cylinder?
An upright cylinder with a height of 30 and a radius of 5 is in a big tub being filled with oil. If only the top 10% of the cylinder is visible, what is the surface area of the submerged cylinder?
The height of the submerged part of the cylinder is 27cm. 2πrh + πr2 is equal to 270π + 25π = 295π
The height of the submerged part of the cylinder is 27cm. 2πrh + πr2 is equal to 270π + 25π = 295π
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A right circular cylinder has a height of 41 in. and a lateral area (excluding top and bottom) 512.5π in2. What is the area of its bases?
A right circular cylinder has a height of 41 in. and a lateral area (excluding top and bottom) 512.5π in2. What is the area of its bases?
The lateral area (not including its bases) is equal to the circumference of the base times the height of the cylinder. Think of it like a label that is wrapped around a soup can. Therefore, we can write this area as:
A = h * π * d or A = h * π * 2r = 2πrh
Now, substituting in our values, we get:
512.5π = 2 * 41*rπ; 512.5π = 82rπ
Solve for r by dividing both sides by 82π:
6.25 = r
From here, we can calculate the area of a base:
A = 6.252π = 39.0625π
NOTE: The question asks for the area of the bases. Therefore, the answer is 2 * 39.0625π or 78.125π in2.
The lateral area (not including its bases) is equal to the circumference of the base times the height of the cylinder. Think of it like a label that is wrapped around a soup can. Therefore, we can write this area as:
A = h * π * d or A = h * π * 2r = 2πrh
Now, substituting in our values, we get:
512.5π = 2 * 41*rπ; 512.5π = 82rπ
Solve for r by dividing both sides by 82π:
6.25 = r
From here, we can calculate the area of a base:
A = 6.252π = 39.0625π
NOTE: The question asks for the area of the bases. Therefore, the answer is 2 * 39.0625π or 78.125π in2.
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The diameter of the lid of a right cylindrical soup can is 5 in. If the can is 12 inches tall and the label costs $0.00125 per square inch to print, what is the cost to produce a label for a can? (Round to the nearest cent.)
The diameter of the lid of a right cylindrical soup can is 5 in. If the can is 12 inches tall and the label costs $0.00125 per square inch to print, what is the cost to produce a label for a can? (Round to the nearest cent.)
The general mechanics of this problem are simple. The lateral area of a right cylinder (excluding its top and bottom) is equal to the circumference of the top times the height of the cylinder. Therefore, the area of this can's surface is: 5π * 12 or 60π. If the cost per square inch is $0.00125, a single label will cost 0.00125 * 60π or $0.075π or approximately $0.24.
The general mechanics of this problem are simple. The lateral area of a right cylinder (excluding its top and bottom) is equal to the circumference of the top times the height of the cylinder. Therefore, the area of this can's surface is: 5π * 12 or 60π. If the cost per square inch is $0.00125, a single label will cost 0.00125 * 60π or $0.075π or approximately $0.24.
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Aluminum is sold to a soup manufacturer at a rate of $0.0015 per square inch. The cans are made so that the ends perfectly fit on the cylindrical body of the can. It costs $0.00125 to attach the ends to the can. The outer label (not covering the top / bottom) costs $0.0001 per in2 to print and stick to the can. The label must be 2 inches longer than circumference of the can. Ignoring any potential waste, what is the manufacturing cost (to the nearest cent) for a can with a radius of 5 inches and a height of 12 inches?
Aluminum is sold to a soup manufacturer at a rate of $0.0015 per square inch. The cans are made so that the ends perfectly fit on the cylindrical body of the can. It costs $0.00125 to attach the ends to the can. The outer label (not covering the top / bottom) costs $0.0001 per in2 to print and stick to the can. The label must be 2 inches longer than circumference of the can. Ignoring any potential waste, what is the manufacturing cost (to the nearest cent) for a can with a radius of 5 inches and a height of 12 inches?
We have the following categories to consider:
= ( + ) * 0.0015
= 2 * 0.00125 = $0.0025
The area of ends of the can are each equal to π*52 or 25π. For two ends, that is 50π.
The lateral area of the can is equal to the circumference of the top times the height, or 2 * π * r * h = 2 * 5 * 12 * π = 120π.
Therefore, the total surface area of the aluminum can is 120π + 50π = 170π. The cost is 170π * 0.0015 = 0.255π, or approximately $0.80.
The area of the label is NOT the same as the lateral area of the can. (Recall that it must be 2 inches longer than the circumference of the can.) Therefore, the area of the label is (2 + 2 * π * 5) * 12 = (2 + 10π) * 12 = 24 + 120π. Multiply this by 0.0001 to get 0.0024 + 0.012π = (approximately) $0.04.
Therefore, the total cost is approximately 0.80 + 0.04 + 0.0025 = $0.8425, or $0.84.
We have the following categories to consider:
The area of ends of the can are each equal to π*52 or 25π. For two ends, that is 50π.
The lateral area of the can is equal to the circumference of the top times the height, or 2 * π * r * h = 2 * 5 * 12 * π = 120π.
Therefore, the total surface area of the aluminum can is 120π + 50π = 170π. The cost is 170π * 0.0015 = 0.255π, or approximately $0.80.
The area of the label is NOT the same as the lateral area of the can. (Recall that it must be 2 inches longer than the circumference of the can.) Therefore, the area of the label is (2 + 2 * π * 5) * 12 = (2 + 10π) * 12 = 24 + 120π. Multiply this by 0.0001 to get 0.0024 + 0.012π = (approximately) $0.04.
Therefore, the total cost is approximately 0.80 + 0.04 + 0.0025 = $0.8425, or $0.84.
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The number of square units in the surface area of a right circular cylinder is equal to the number of cubic units in its volume. If r and h represent the length in units of the cylinder's radius and height, respectively, which of the following is equivalent to r in terms of h?
The number of square units in the surface area of a right circular cylinder is equal to the number of cubic units in its volume. If r and h represent the length in units of the cylinder's radius and height, respectively, which of the following is equivalent to r in terms of h?
We need to find expressions for the surface area and the volume of a cylinder. The surface area of the cylinder consists of the sum of the surface areas of the two bases plus the lateral surface area.
surface area of cylinder = surface area of bases + lateral surface area
The bases of the cylinder will be two circles with radius r. Thus, the area of each will be _πr_2, and their combined surface area will be 2_πr_2.
The lateral surface area of the cylinder is equal to the circumference of the circular base multiplied by the height. The circumferece of a circle is 2_πr_, and the height is h, so the lateral area is 2_πrh_.
surface area of cylinder = 2_πr_2 + 2_πrh_
Next, we need to find an expression for the volume. The volume of a cylinder is equal to the product of the height and the area of one of the bases. The area of the base is _πr_2, and the height is h, so the volume of the cylinder is πr_2_h.
volume = πr_2_h
Then, we must set the volume and surface area expressions equal to one another and solve for r in terms of h.
2_πr_2 + 2_πrh_ = πr_2_h
First, let's factor out 2_πr_ from the left side.
2_πr_(r + h) = πr_2_h
We can divide both sides by π.
2_r_(r + h) = r_2_h
We can also divide both sides by r, because the radius cannot equal zero.
2(r + h) = rh
Let's now distribute the 2 on the left side.
2_r_ + 2_h_ = rh
Subtract 2_r_ from both sides to get all the r's on one side.
2_h_ = rh – 2_r_
r_h – 2_r = 2_h_
Factor out an r from the left side.
r(h – 2) = 2_h_
Divide both sides by h – 2
r = 2_h_/(h – 2)
The answer is r = 2_h_/(h – 2).
We need to find expressions for the surface area and the volume of a cylinder. The surface area of the cylinder consists of the sum of the surface areas of the two bases plus the lateral surface area.
surface area of cylinder = surface area of bases + lateral surface area
The bases of the cylinder will be two circles with radius r. Thus, the area of each will be _πr_2, and their combined surface area will be 2_πr_2.
The lateral surface area of the cylinder is equal to the circumference of the circular base multiplied by the height. The circumferece of a circle is 2_πr_, and the height is h, so the lateral area is 2_πrh_.
surface area of cylinder = 2_πr_2 + 2_πrh_
Next, we need to find an expression for the volume. The volume of a cylinder is equal to the product of the height and the area of one of the bases. The area of the base is _πr_2, and the height is h, so the volume of the cylinder is πr_2_h.
volume = πr_2_h
Then, we must set the volume and surface area expressions equal to one another and solve for r in terms of h.
2_πr_2 + 2_πrh_ = πr_2_h
First, let's factor out 2_πr_ from the left side.
2_πr_(r + h) = πr_2_h
We can divide both sides by π.
2_r_(r + h) = r_2_h
We can also divide both sides by r, because the radius cannot equal zero.
2(r + h) = rh
Let's now distribute the 2 on the left side.
2_r_ + 2_h_ = rh
Subtract 2_r_ from both sides to get all the r's on one side.
2_h_ = rh – 2_r_
r_h – 2_r = 2_h_
Factor out an r from the left side.
r(h – 2) = 2_h_
Divide both sides by h – 2
r = 2_h_/(h – 2)
The answer is r = 2_h_/(h – 2).
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What is the surface area of a cylinder with a radius of 
and a height of 
?
What is the surface area of a cylinder with a radius of 
When you're calculating the surface area of a cylinder, note that the cylinder will have two circles, one for the top and one for the bottom, and one rectangle that wraps around the "side" of the cylinder (it's helpful to picture peeling the label off a can of soup - it's curved when it's on the can, but really it's a rectangle that has been wrapped around). You know the area of the circle formula; for the rectangle, note that the height is given to you but the width of the rectangle is one you have to intuit: it's the circumference of the circle, because the entire distance around the circle from one point around and back again is the horizontal distance that the area must cover.
Therefore the surface area of a cylinder = 
When you're calculating the surface area of a cylinder, note that the cylinder will have two circles, one for the top and one for the bottom, and one rectangle that wraps around the "side" of the cylinder (it's helpful to picture peeling the label off a can of soup - it's curved when it's on the can, but really it's a rectangle that has been wrapped around). You know the area of the circle formula; for the rectangle, note that the height is given to you but the width of the rectangle is one you have to intuit: it's the circumference of the circle, because the entire distance around the circle from one point around and back again is the horizontal distance that the area must cover.
Therefore the surface area of a cylinder =
Compare your answer with the correct one above
An upright cylinder with a height of 30 and a radius of 5 is in a big tub being filled with oil. If only the top 10% of the cylinder is visible, what is the surface area of the submerged cylinder?
An upright cylinder with a height of 30 and a radius of 5 is in a big tub being filled with oil. If only the top 10% of the cylinder is visible, what is the surface area of the submerged cylinder?
The height of the submerged part of the cylinder is 27cm. 2πrh + πr2 is equal to 270π + 25π = 295π
The height of the submerged part of the cylinder is 27cm. 2πrh + πr2 is equal to 270π + 25π = 295π
Compare your answer with the correct one above
A right circular cylinder has a height of 41 in. and a lateral area (excluding top and bottom) 512.5π in2. What is the area of its bases?
A right circular cylinder has a height of 41 in. and a lateral area (excluding top and bottom) 512.5π in2. What is the area of its bases?
The lateral area (not including its bases) is equal to the circumference of the base times the height of the cylinder. Think of it like a label that is wrapped around a soup can. Therefore, we can write this area as:
A = h * π * d or A = h * π * 2r = 2πrh
Now, substituting in our values, we get:
512.5π = 2 * 41*rπ; 512.5π = 82rπ
Solve for r by dividing both sides by 82π:
6.25 = r
From here, we can calculate the area of a base:
A = 6.252π = 39.0625π
NOTE: The question asks for the area of the bases. Therefore, the answer is 2 * 39.0625π or 78.125π in2.
The lateral area (not including its bases) is equal to the circumference of the base times the height of the cylinder. Think of it like a label that is wrapped around a soup can. Therefore, we can write this area as:
A = h * π * d or A = h * π * 2r = 2πrh
Now, substituting in our values, we get:
512.5π = 2 * 41*rπ; 512.5π = 82rπ
Solve for r by dividing both sides by 82π:
6.25 = r
From here, we can calculate the area of a base:
A = 6.252π = 39.0625π
NOTE: The question asks for the area of the bases. Therefore, the answer is 2 * 39.0625π or 78.125π in2.
Compare your answer with the correct one above
The diameter of the lid of a right cylindrical soup can is 5 in. If the can is 12 inches tall and the label costs $0.00125 per square inch to print, what is the cost to produce a label for a can? (Round to the nearest cent.)
The diameter of the lid of a right cylindrical soup can is 5 in. If the can is 12 inches tall and the label costs $0.00125 per square inch to print, what is the cost to produce a label for a can? (Round to the nearest cent.)
The general mechanics of this problem are simple. The lateral area of a right cylinder (excluding its top and bottom) is equal to the circumference of the top times the height of the cylinder. Therefore, the area of this can's surface is: 5π * 12 or 60π. If the cost per square inch is $0.00125, a single label will cost 0.00125 * 60π or $0.075π or approximately $0.24.
The general mechanics of this problem are simple. The lateral area of a right cylinder (excluding its top and bottom) is equal to the circumference of the top times the height of the cylinder. Therefore, the area of this can's surface is: 5π * 12 or 60π. If the cost per square inch is $0.00125, a single label will cost 0.00125 * 60π or $0.075π or approximately $0.24.
Compare your answer with the correct one above
Aluminum is sold to a soup manufacturer at a rate of $0.0015 per square inch. The cans are made so that the ends perfectly fit on the cylindrical body of the can. It costs $0.00125 to attach the ends to the can. The outer label (not covering the top / bottom) costs $0.0001 per in2 to print and stick to the can. The label must be 2 inches longer than circumference of the can. Ignoring any potential waste, what is the manufacturing cost (to the nearest cent) for a can with a radius of 5 inches and a height of 12 inches?
Aluminum is sold to a soup manufacturer at a rate of $0.0015 per square inch. The cans are made so that the ends perfectly fit on the cylindrical body of the can. It costs $0.00125 to attach the ends to the can. The outer label (not covering the top / bottom) costs $0.0001 per in2 to print and stick to the can. The label must be 2 inches longer than circumference of the can. Ignoring any potential waste, what is the manufacturing cost (to the nearest cent) for a can with a radius of 5 inches and a height of 12 inches?
We have the following categories to consider:
= ( + ) * 0.0015
= 2 * 0.00125 = $0.0025
The area of ends of the can are each equal to π*52 or 25π. For two ends, that is 50π.
The lateral area of the can is equal to the circumference of the top times the height, or 2 * π * r * h = 2 * 5 * 12 * π = 120π.
Therefore, the total surface area of the aluminum can is 120π + 50π = 170π. The cost is 170π * 0.0015 = 0.255π, or approximately $0.80.
The area of the label is NOT the same as the lateral area of the can. (Recall that it must be 2 inches longer than the circumference of the can.) Therefore, the area of the label is (2 + 2 * π * 5) * 12 = (2 + 10π) * 12 = 24 + 120π. Multiply this by 0.0001 to get 0.0024 + 0.012π = (approximately) $0.04.
Therefore, the total cost is approximately 0.80 + 0.04 + 0.0025 = $0.8425, or $0.84.
We have the following categories to consider:
The area of ends of the can are each equal to π*52 or 25π. For two ends, that is 50π.
The lateral area of the can is equal to the circumference of the top times the height, or 2 * π * r * h = 2 * 5 * 12 * π = 120π.
Therefore, the total surface area of the aluminum can is 120π + 50π = 170π. The cost is 170π * 0.0015 = 0.255π, or approximately $0.80.
The area of the label is NOT the same as the lateral area of the can. (Recall that it must be 2 inches longer than the circumference of the can.) Therefore, the area of the label is (2 + 2 * π * 5) * 12 = (2 + 10π) * 12 = 24 + 120π. Multiply this by 0.0001 to get 0.0024 + 0.012π = (approximately) $0.04.
Therefore, the total cost is approximately 0.80 + 0.04 + 0.0025 = $0.8425, or $0.84.
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The number of square units in the surface area of a right circular cylinder is equal to the number of cubic units in its volume. If r and h represent the length in units of the cylinder's radius and height, respectively, which of the following is equivalent to r in terms of h?
The number of square units in the surface area of a right circular cylinder is equal to the number of cubic units in its volume. If r and h represent the length in units of the cylinder's radius and height, respectively, which of the following is equivalent to r in terms of h?
We need to find expressions for the surface area and the volume of a cylinder. The surface area of the cylinder consists of the sum of the surface areas of the two bases plus the lateral surface area.
surface area of cylinder = surface area of bases + lateral surface area
The bases of the cylinder will be two circles with radius r. Thus, the area of each will be _πr_2, and their combined surface area will be 2_πr_2.
The lateral surface area of the cylinder is equal to the circumference of the circular base multiplied by the height. The circumferece of a circle is 2_πr_, and the height is h, so the lateral area is 2_πrh_.
surface area of cylinder = 2_πr_2 + 2_πrh_
Next, we need to find an expression for the volume. The volume of a cylinder is equal to the product of the height and the area of one of the bases. The area of the base is _πr_2, and the height is h, so the volume of the cylinder is πr_2_h.
volume = πr_2_h
Then, we must set the volume and surface area expressions equal to one another and solve for r in terms of h.
2_πr_2 + 2_πrh_ = πr_2_h
First, let's factor out 2_πr_ from the left side.
2_πr_(r + h) = πr_2_h
We can divide both sides by π.
2_r_(r + h) = r_2_h
We can also divide both sides by r, because the radius cannot equal zero.
2(r + h) = rh
Let's now distribute the 2 on the left side.
2_r_ + 2_h_ = rh
Subtract 2_r_ from both sides to get all the r's on one side.
2_h_ = rh – 2_r_
r_h – 2_r = 2_h_
Factor out an r from the left side.
r(h – 2) = 2_h_
Divide both sides by h – 2
r = 2_h_/(h – 2)
The answer is r = 2_h_/(h – 2).
We need to find expressions for the surface area and the volume of a cylinder. The surface area of the cylinder consists of the sum of the surface areas of the two bases plus the lateral surface area.
surface area of cylinder = surface area of bases + lateral surface area
The bases of the cylinder will be two circles with radius r. Thus, the area of each will be _πr_2, and their combined surface area will be 2_πr_2.
The lateral surface area of the cylinder is equal to the circumference of the circular base multiplied by the height. The circumferece of a circle is 2_πr_, and the height is h, so the lateral area is 2_πrh_.
surface area of cylinder = 2_πr_2 + 2_πrh_
Next, we need to find an expression for the volume. The volume of a cylinder is equal to the product of the height and the area of one of the bases. The area of the base is _πr_2, and the height is h, so the volume of the cylinder is πr_2_h.
volume = πr_2_h
Then, we must set the volume and surface area expressions equal to one another and solve for r in terms of h.
2_πr_2 + 2_πrh_ = πr_2_h
First, let's factor out 2_πr_ from the left side.
2_πr_(r + h) = πr_2_h
We can divide both sides by π.
2_r_(r + h) = r_2_h
We can also divide both sides by r, because the radius cannot equal zero.
2(r + h) = rh
Let's now distribute the 2 on the left side.
2_r_ + 2_h_ = rh
Subtract 2_r_ from both sides to get all the r's on one side.
2_h_ = rh – 2_r_
r_h – 2_r = 2_h_
Factor out an r from the left side.
r(h – 2) = 2_h_
Divide both sides by h – 2
r = 2_h_/(h – 2)
The answer is r = 2_h_/(h – 2).
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What is the surface area of a cylinder with a radius of and a height of ?
What is the surface area of a cylinder with a radius of
When you're calculating the surface area of a cylinder, note that the cylinder will have two circles, one for the top and one for the bottom, and one rectangle that wraps around the "side" of the cylinder (it's helpful to picture peeling the label off a can of soup - it's curved when it's on the can, but really it's a rectangle that has been wrapped around). You know the area of the circle formula; for the rectangle, note that the height is given to you but the width of the rectangle is one you have to intuit: it's the circumference of the circle, because the entire distance around the circle from one point around and back again is the horizontal distance that the area must cover.
Therefore the surface area of a cylinder =
When you're calculating the surface area of a cylinder, note that the cylinder will have two circles, one for the top and one for the bottom, and one rectangle that wraps around the "side" of the cylinder (it's helpful to picture peeling the label off a can of soup - it's curved when it's on the can, but really it's a rectangle that has been wrapped around). You know the area of the circle formula; for the rectangle, note that the height is given to you but the width of the rectangle is one you have to intuit: it's the circumference of the circle, because the entire distance around the circle from one point around and back again is the horizontal distance that the area must cover.
Therefore the surface area of a cylinder =
Compare your answer with the correct one above
An upright cylinder with a height of 30 and a radius of 5 is in a big tub being filled with oil. If only the top 10% of the cylinder is visible, what is the surface area of the submerged cylinder?
An upright cylinder with a height of 30 and a radius of 5 is in a big tub being filled with oil. If only the top 10% of the cylinder is visible, what is the surface area of the submerged cylinder?
The height of the submerged part of the cylinder is 27cm. 2πrh + πr2 is equal to 270π + 25π = 295π
The height of the submerged part of the cylinder is 27cm. 2πrh + πr2 is equal to 270π + 25π = 295π
Compare your answer with the correct one above
A right circular cylinder has a height of 41 in. and a lateral area (excluding top and bottom) 512.5π in2. What is the area of its bases?
A right circular cylinder has a height of 41 in. and a lateral area (excluding top and bottom) 512.5π in2. What is the area of its bases?
The lateral area (not including its bases) is equal to the circumference of the base times the height of the cylinder. Think of it like a label that is wrapped around a soup can. Therefore, we can write this area as:
A = h * π * d or A = h * π * 2r = 2πrh
Now, substituting in our values, we get:
512.5π = 2 * 41*rπ; 512.5π = 82rπ
Solve for r by dividing both sides by 82π:
6.25 = r
From here, we can calculate the area of a base:
A = 6.252π = 39.0625π
NOTE: The question asks for the area of the bases. Therefore, the answer is 2 * 39.0625π or 78.125π in2.
The lateral area (not including its bases) is equal to the circumference of the base times the height of the cylinder. Think of it like a label that is wrapped around a soup can. Therefore, we can write this area as:
A = h * π * d or A = h * π * 2r = 2πrh
Now, substituting in our values, we get:
512.5π = 2 * 41*rπ; 512.5π = 82rπ
Solve for r by dividing both sides by 82π:
6.25 = r
From here, we can calculate the area of a base:
A = 6.252π = 39.0625π
NOTE: The question asks for the area of the bases. Therefore, the answer is 2 * 39.0625π or 78.125π in2.
Compare your answer with the correct one above
The diameter of the lid of a right cylindrical soup can is 5 in. If the can is 12 inches tall and the label costs $0.00125 per square inch to print, what is the cost to produce a label for a can? (Round to the nearest cent.)
The diameter of the lid of a right cylindrical soup can is 5 in. If the can is 12 inches tall and the label costs $0.00125 per square inch to print, what is the cost to produce a label for a can? (Round to the nearest cent.)
The general mechanics of this problem are simple. The lateral area of a right cylinder (excluding its top and bottom) is equal to the circumference of the top times the height of the cylinder. Therefore, the area of this can's surface is: 5π * 12 or 60π. If the cost per square inch is $0.00125, a single label will cost 0.00125 * 60π or $0.075π or approximately $0.24.
The general mechanics of this problem are simple. The lateral area of a right cylinder (excluding its top and bottom) is equal to the circumference of the top times the height of the cylinder. Therefore, the area of this can's surface is: 5π * 12 or 60π. If the cost per square inch is $0.00125, a single label will cost 0.00125 * 60π or $0.075π or approximately $0.24.
Compare your answer with the correct one above
Aluminum is sold to a soup manufacturer at a rate of $0.0015 per square inch. The cans are made so that the ends perfectly fit on the cylindrical body of the can. It costs $0.00125 to attach the ends to the can. The outer label (not covering the top / bottom) costs $0.0001 per in2 to print and stick to the can. The label must be 2 inches longer than circumference of the can. Ignoring any potential waste, what is the manufacturing cost (to the nearest cent) for a can with a radius of 5 inches and a height of 12 inches?
Aluminum is sold to a soup manufacturer at a rate of $0.0015 per square inch. The cans are made so that the ends perfectly fit on the cylindrical body of the can. It costs $0.00125 to attach the ends to the can. The outer label (not covering the top / bottom) costs $0.0001 per in2 to print and stick to the can. The label must be 2 inches longer than circumference of the can. Ignoring any potential waste, what is the manufacturing cost (to the nearest cent) for a can with a radius of 5 inches and a height of 12 inches?
We have the following categories to consider:
= ( + ) * 0.0015
= 2 * 0.00125 = $0.0025
The area of ends of the can are each equal to π*52 or 25π. For two ends, that is 50π.
The lateral area of the can is equal to the circumference of the top times the height, or 2 * π * r * h = 2 * 5 * 12 * π = 120π.
Therefore, the total surface area of the aluminum can is 120π + 50π = 170π. The cost is 170π * 0.0015 = 0.255π, or approximately $0.80.
The area of the label is NOT the same as the lateral area of the can. (Recall that it must be 2 inches longer than the circumference of the can.) Therefore, the area of the label is (2 + 2 * π * 5) * 12 = (2 + 10π) * 12 = 24 + 120π. Multiply this by 0.0001 to get 0.0024 + 0.012π = (approximately) $0.04.
Therefore, the total cost is approximately 0.80 + 0.04 + 0.0025 = $0.8425, or $0.84.
We have the following categories to consider:
The area of ends of the can are each equal to π*52 or 25π. For two ends, that is 50π.
The lateral area of the can is equal to the circumference of the top times the height, or 2 * π * r * h = 2 * 5 * 12 * π = 120π.
Therefore, the total surface area of the aluminum can is 120π + 50π = 170π. The cost is 170π * 0.0015 = 0.255π, or approximately $0.80.
The area of the label is NOT the same as the lateral area of the can. (Recall that it must be 2 inches longer than the circumference of the can.) Therefore, the area of the label is (2 + 2 * π * 5) * 12 = (2 + 10π) * 12 = 24 + 120π. Multiply this by 0.0001 to get 0.0024 + 0.012π = (approximately) $0.04.
Therefore, the total cost is approximately 0.80 + 0.04 + 0.0025 = $0.8425, or $0.84.
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The number of square units in the surface area of a right circular cylinder is equal to the number of cubic units in its volume. If r and h represent the length in units of the cylinder's radius and height, respectively, which of the following is equivalent to r in terms of h?
The number of square units in the surface area of a right circular cylinder is equal to the number of cubic units in its volume. If r and h represent the length in units of the cylinder's radius and height, respectively, which of the following is equivalent to r in terms of h?
We need to find expressions for the surface area and the volume of a cylinder. The surface area of the cylinder consists of the sum of the surface areas of the two bases plus the lateral surface area.
surface area of cylinder = surface area of bases + lateral surface area
The bases of the cylinder will be two circles with radius r. Thus, the area of each will be _πr_2, and their combined surface area will be 2_πr_2.
The lateral surface area of the cylinder is equal to the circumference of the circular base multiplied by the height. The circumferece of a circle is 2_πr_, and the height is h, so the lateral area is 2_πrh_.
surface area of cylinder = 2_πr_2 + 2_πrh_
Next, we need to find an expression for the volume. The volume of a cylinder is equal to the product of the height and the area of one of the bases. The area of the base is _πr_2, and the height is h, so the volume of the cylinder is πr_2_h.
volume = πr_2_h
Then, we must set the volume and surface area expressions equal to one another and solve for r in terms of h.
2_πr_2 + 2_πrh_ = πr_2_h
First, let's factor out 2_πr_ from the left side.
2_πr_(r + h) = πr_2_h
We can divide both sides by π.
2_r_(r + h) = r_2_h
We can also divide both sides by r, because the radius cannot equal zero.
2(r + h) = rh
Let's now distribute the 2 on the left side.
2_r_ + 2_h_ = rh
Subtract 2_r_ from both sides to get all the r's on one side.
2_h_ = rh – 2_r_
r_h – 2_r = 2_h_
Factor out an r from the left side.
r(h – 2) = 2_h_
Divide both sides by h – 2
r = 2_h_/(h – 2)
The answer is r = 2_h_/(h – 2).
We need to find expressions for the surface area and the volume of a cylinder. The surface area of the cylinder consists of the sum of the surface areas of the two bases plus the lateral surface area.
surface area of cylinder = surface area of bases + lateral surface area
The bases of the cylinder will be two circles with radius r. Thus, the area of each will be _πr_2, and their combined surface area will be 2_πr_2.
The lateral surface area of the cylinder is equal to the circumference of the circular base multiplied by the height. The circumferece of a circle is 2_πr_, and the height is h, so the lateral area is 2_πrh_.
surface area of cylinder = 2_πr_2 + 2_πrh_
Next, we need to find an expression for the volume. The volume of a cylinder is equal to the product of the height and the area of one of the bases. The area of the base is _πr_2, and the height is h, so the volume of the cylinder is πr_2_h.
volume = πr_2_h
Then, we must set the volume and surface area expressions equal to one another and solve for r in terms of h.
2_πr_2 + 2_πrh_ = πr_2_h
First, let's factor out 2_πr_ from the left side.
2_πr_(r + h) = πr_2_h
We can divide both sides by π.
2_r_(r + h) = r_2_h
We can also divide both sides by r, because the radius cannot equal zero.
2(r + h) = rh
Let's now distribute the 2 on the left side.
2_r_ + 2_h_ = rh
Subtract 2_r_ from both sides to get all the r's on one side.
2_h_ = rh – 2_r_
r_h – 2_r = 2_h_
Factor out an r from the left side.
r(h – 2) = 2_h_
Divide both sides by h – 2
r = 2_h_/(h – 2)
The answer is r = 2_h_/(h – 2).
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What is the surface area of a cylinder with a radius of and a height of ?
What is the surface area of a cylinder with a radius of
When you're calculating the surface area of a cylinder, note that the cylinder will have two circles, one for the top and one for the bottom, and one rectangle that wraps around the "side" of the cylinder (it's helpful to picture peeling the label off a can of soup - it's curved when it's on the can, but really it's a rectangle that has been wrapped around). You know the area of the circle formula; for the rectangle, note that the height is given to you but the width of the rectangle is one you have to intuit: it's the circumference of the circle, because the entire distance around the circle from one point around and back again is the horizontal distance that the area must cover.
Therefore the surface area of a cylinder =
When you're calculating the surface area of a cylinder, note that the cylinder will have two circles, one for the top and one for the bottom, and one rectangle that wraps around the "side" of the cylinder (it's helpful to picture peeling the label off a can of soup - it's curved when it's on the can, but really it's a rectangle that has been wrapped around). You know the area of the circle formula; for the rectangle, note that the height is given to you but the width of the rectangle is one you have to intuit: it's the circumference of the circle, because the entire distance around the circle from one point around and back again is the horizontal distance that the area must cover.
Therefore the surface area of a cylinder =
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An upright cylinder with a height of 30 and a radius of 5 is in a big tub being filled with oil. If only the top 10% of the cylinder is visible, what is the surface area of the submerged cylinder?
An upright cylinder with a height of 30 and a radius of 5 is in a big tub being filled with oil. If only the top 10% of the cylinder is visible, what is the surface area of the submerged cylinder?
The height of the submerged part of the cylinder is 27cm. 2πrh + πr2 is equal to 270π + 25π = 295π
The height of the submerged part of the cylinder is 27cm. 2πrh + πr2 is equal to 270π + 25π = 295π
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A right circular cylinder has a height of 41 in. and a lateral area (excluding top and bottom) 512.5π in2. What is the area of its bases?
A right circular cylinder has a height of 41 in. and a lateral area (excluding top and bottom) 512.5π in2. What is the area of its bases?
The lateral area (not including its bases) is equal to the circumference of the base times the height of the cylinder. Think of it like a label that is wrapped around a soup can. Therefore, we can write this area as:
A = h * π * d or A = h * π * 2r = 2πrh
Now, substituting in our values, we get:
512.5π = 2 * 41*rπ; 512.5π = 82rπ
Solve for r by dividing both sides by 82π:
6.25 = r
From here, we can calculate the area of a base:
A = 6.252π = 39.0625π
NOTE: The question asks for the area of the bases. Therefore, the answer is 2 * 39.0625π or 78.125π in2.
The lateral area (not including its bases) is equal to the circumference of the base times the height of the cylinder. Think of it like a label that is wrapped around a soup can. Therefore, we can write this area as:
A = h * π * d or A = h * π * 2r = 2πrh
Now, substituting in our values, we get:
512.5π = 2 * 41*rπ; 512.5π = 82rπ
Solve for r by dividing both sides by 82π:
6.25 = r
From here, we can calculate the area of a base:
A = 6.252π = 39.0625π
NOTE: The question asks for the area of the bases. Therefore, the answer is 2 * 39.0625π or 78.125π in2.
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