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Calculus 3 : Spherical Coordinates

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Example Questions

Example Question #1 : Spherical Coordinates

Convert the following into spherical coordinates.

$x=\sin(t)$

$y=\cos(t)$

z=1

Possible Answers:

$\rho=\sqrt{2}$

$\phi=0$

$\theta=\tan(\cot(t))$

$\rho=2$

$\phi=-\frac{\pi}{4}$

$\theta=\tan^{-1}(\cot(t))$

$\rho=\sqrt{2}$

$\phi=\frac{\pi}{4}$

$\theta=\tan^{-1}(\cot(t))$

$\rho=\sqrt{2}$

$\phi=\frac{\pi}{4}$

$\theta=\tan^{-1}(\cot(t))$

$\rho=1$

$\phi=\frac{\pi}{2}$

$\theta=\tan(\cot(t))$

Correct answer:

$\rho=\sqrt{2}$

$\phi=\frac{\pi}{4}$

$\theta=\tan^{-1}(\cot(t))$

Explanation:

In order to convert to spherical coordinates , we need to remember the conversion equations.

$\rho=\sqrt{x^2+y^2+z^2}$

$\theta=\tan^{-1}\Big(\frac{y}{x}\Big)$

$\phi=\tan^{-1}\Big(\frac{\sqrt{x^2+y^2}}{z}\Big)$

Now lets apply this to our problem.

$\rho=\sqrt{(\sin(t))^2+(\cos(t))^2+1^2}=\sqrt{1+1}=\sqrt{2}$

$\phi=\tan^{-1}\Big(\frac{\sqrt{(\sin{t})^2+(\cos{t})^2}}{1}\Big)=\tan^{-1}(1)=\frac{\pi}{4}$

$\theta=\tan^{-1}\Big(\frac{\cos(t)}{\sin(t)}\Big)=\tan^{-1}(\cot(t))$

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Example Question #2 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (9,3,2) . What is the position of this point in spherical coordinates?

Possible Answers:

$(9.49,18.43^{\circ},78.1^{\circ})$

$(9.49,161.57^{\circ},78.1^{\circ})$

$(9.7,18.43^{\circ},78.1^{\circ})$

$(9.7,161.57^{\circ},78.1^{\circ})$

$(9.49,161.57^{\circ},-101.9^{\circ})$

Correct answer:

$(9.7,18.43^{\circ},78.1^{\circ})$

Explanation:

When given Cartesian coordinates of the form (x,y,z) to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(9,3,2)\\&\rho=\sqrt{(9)^2+(3)^2+(2)^2}=9.7\\&\theta=arctan(\frac{3}{9})=18.43^{\circ}\\&\phi=arccos(\frac{2}{9.7})=78.1^{\circ}\end{align*}$

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Example Question #3 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-12,4,-3) . What is the position of this point in spherical coordinates?

Possible Answers:

$(12.65,18.43^{\circ},-76.66^{\circ})$

$(12.65,161.57^{\circ},103.34^{\circ})$

$(12.65,18.43^{\circ},76.66^{\circ})$

$(13,18.43^{\circ},103.34^{\circ})$

$(13,161.57^{\circ},103.34^{\circ})$

Correct answer:

$(13,161.57^{\circ},103.34^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(-12,4,-3)\\&\rho=\sqrt{(-12)^2+(4)^2+(-3)^2}=13\\&\theta=arctan(\frac{4}{-12})=161.57^{\circ}\\&\phi=arccos(\frac{-3}{13})=103.34^{\circ}\end{align*}$

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Example Question #4 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (5,1,-6) . What is the position of this point in spherical coordinates?

Possible Answers:

$(5.1,168.69^{\circ},40.36^{\circ})$

$(7.87,168.69^{\circ},139.64^{\circ})$

$(7.87,11.31^{\circ},139.64^{\circ})$

$(5.1,168.69^{\circ},-40.36^{\circ})$

$(5.1,11.31^{\circ},139.64^{\circ})$

Correct answer:

$(7.87,11.31^{\circ},139.64^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(5,1,-6)\\&\rho=\sqrt{(5)^2+(1)^2+(-6)^2}=7.87\\&\theta=arctan(\frac{1}{5})=11.31^{\circ}\\&\phi=arccos(\frac{-6}{7.87})=139.64^{\circ}\end{align*}$

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Example Question #2011 : Calculus 3

A point in space is located, in Cartesian coordinates, at (5,-8,3) . What is the position of this point in spherical coordinates?

Possible Answers:

$(9.9,122.01^{\circ},72.36^{\circ})$

$(9.9,-57.99^{\circ},72.36^{\circ})$

$(9.43,-57.99^{\circ},72.36^{\circ})$

$(9.43,122.01^{\circ},107.64^{\circ})$

$(9.43,122.01^{\circ},-107.64^{\circ})$

Correct answer:

$(9.9,-57.99^{\circ},72.36^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(5,-8,3)\\&\rho=\sqrt{(5)^2+(-8)^2+(3)^2}=9.9\\&\theta=arctan(\frac{-8}{5})=-57.99^{\circ}\\&\phi=arccos(\frac{3}{9.9})=72.36^{\circ}\end{align*}$

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Example Question #6 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-9,-9,-9) . What is the position of this point in spherical coordinates?

Possible Answers:

$(12.73,45^{\circ},54.74^{\circ})$

$(15.59,-135^{\circ},125.26^{\circ})$

$(15.59,45^{\circ},125.26^{\circ})$

$(12.73,45^{\circ},-54.74^{\circ})$

$(12.73,-135^{\circ},125.26^{\circ})$

Correct answer:

$(15.59,-135^{\circ},125.26^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(-9,-9,-9)\\&\rho=\sqrt{(-9)^2+(-9)^2+(-9)^2}=15.59\\&\theta=arctan(\frac{-9}{-9})=-135^{\circ}\\&\phi=arccos(\frac{-9}{15.59})=125.26^{\circ}\end{align*}$

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Example Question #2011 : Calculus 3

A point in space is located, in Cartesian coordinates, at (11,0,-5) . What is the position of this point in spherical coordinates?

Possible Answers:

$(11,0^{\circ},114.44^{\circ})$

$(11,180^{\circ},-65.56^{\circ})$

$(12.08,180^{\circ},114.44^{\circ})$

$(11,180^{\circ},65.56^{\circ})$

$(12.08,0^{\circ},114.44^{\circ})$

Correct answer:

$(12.08,0^{\circ},114.44^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(11,0,-5)\\&\rho=\sqrt{(11)^2+(0)^2+(-5)^2}=12.08\\&\theta=arctan(\frac{0}{11})=0^{\circ}\\&\phi=arccos(\frac{-5}{12.08})=114.44^{\circ}\end{align*}$

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Example Question #8 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (85,10,13) . What is the position of this point in spherical coordinates?

Possible Answers:

$(86.57,6.71^{\circ},81.36^{\circ})$

$(85.59,173.29^{\circ},98.64^{\circ})$

$(85.59,173.29^{\circ},-98.64^{\circ})$

$(86.57,173.29^{\circ},81.36^{\circ})$

$(85.59,6.71^{\circ},81.36^{\circ})$

Correct answer:

$(86.57,6.71^{\circ},81.36^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(85,10,13)\\&\rho=\sqrt{(85)^2+(10)^2+(13)^2}=86.57\\&\theta=arctan(\frac{10}{85})=6.71^{\circ}\\&\phi=arccos(\frac{13}{86.57})=81.36^{\circ}\end{align*}$

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Example Question #1 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (3,-23,50) . What is the position of this point in spherical coordinates?

Possible Answers:

$(23.19,97.43^{\circ},155.11^{\circ})$

$(55.12,-82.57^{\circ},24.89^{\circ})$

$(23.19,-82.57^{\circ},24.89^{\circ})$

$(55.12,97.43^{\circ},24.89^{\circ})$

$(23.19,97.43^{\circ},-155.11^{\circ})$

Correct answer:

$(55.12,-82.57^{\circ},24.89^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(3,-23,50)\\&\rho=\sqrt{(3)^2+(-23)^2+(50)^2}=55.12\\&\theta=arctan(\frac{-23}{3})=-82.57^{\circ}\\&\phi=arccos(\frac{50}{55.12})=24.89^{\circ}\end{align*}$

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Example Question #10 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (11,12,13) . What is the position of this point in spherical coordinates?

Possible Answers:

$(20.83,47.49^{\circ},51.39^{\circ})$

$(16.28,132.51^{\circ},-128.61^{\circ})$

$(16.28,47.49^{\circ},51.39^{\circ})$

$(20.83,132.51^{\circ},51.39^{\circ})$

$(16.28,132.51^{\circ},128.61^{\circ})$

Correct answer:

$(20.83,47.49^{\circ},51.39^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(11,12,13)\\&\rho=\sqrt{(11)^2+(12)^2+(13)^2}=20.83\\&\theta=arctan(\frac{12}{11})=47.49^{\circ}\\&\phi=arccos(\frac{13}{20.83})=51.39^{\circ}\end{align*}$

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Calculus 3 : Spherical Coordinates

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Spherical Coordinates

Example Question #2 : Spherical Coordinates

Example Question #3 : Spherical Coordinates

Example Question #4 : Spherical Coordinates

Example Question #2011 : Calculus 3

Example Question #6 : Spherical Coordinates

Example Question #2011 : Calculus 3

Example Question #8 : Spherical Coordinates

Example Question #1 : Spherical Coordinates

Example Question #10 : Spherical Coordinates

All Calculus 3 Resources

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