First, it is important to consider the shape of this solid. This solid is a pyramid, with one square face and four triangular faces. Through a relationship of similar triangles, we are able to relate the known information (a height of $\displaystyle 10$ and a base side length of $\displaystyle 3$) to our general variables for side length $\displaystyle (x)$ and height of the pyramid $\displaystyle (y)$. We can think of this plotted on the coordinate plane, with the width of the pyramid solid being in the $\displaystyle z$ direction.

$\displaystyle \frac{x}{3}=\frac{y}{10} \Rightarrow x=\frac{3y}{10}$

Since the side length $\displaystyle x$ can be squared to find a general formula for the area of the pyramid’s base $\displaystyle (A=x^2)$, this can be applied to the volume using cross-sections formula $\displaystyle (V=\int_{a}^{b}A(x) dx)$ as the next step. Note that the equation $\displaystyle x=\frac{3y}{10}$ solved for above is in terms of $\displaystyle y$. Our new volume function, therefore, is also in terms of $\displaystyle y: V=\int_{a}^{b}(\frac{3y}{10})^2 \: dy$

Because the pyramid reaches a maximum height of $\displaystyle 10$, and we assume the pyramid’s starting height is at $\displaystyle y=0$, the appropriate bounds for the integral expression are: $\displaystyle 0\leq y\leq 10$.

To wrap up this problem, combine the above information into one cohesive expression:

$\displaystyle V=\int_{0}^{10}(\frac{3y}{10})^2 dy =\frac{9}{100} \int_{0}^{10}y^2dy =\frac{9}{100}[\frac{1}{3}y^3] |_0^{10}=\frac{9}{100}*\frac{1}{3}*1000=30$

The cross-sections are perpendicular to the $\displaystyle x$ axis; therefore, the volume expression will be in terms of $\displaystyle x$. 

The area of a rectangle is $\displaystyle A=l*w$. By applying this formula to our general volume formula $\displaystyle (V=\int_{a}^{b}A(x) dx)$, we get the following: $\displaystyle V=\int_{a}^{b}lw \: dx$.

Next, an expression for $\displaystyle lw$ must be determined. The problem specifies the length (or height) of the rectangle cross-sections is $\displaystyle 3$. This just leaves the value of w to be found. The width of the rectangle will vary as the region creating the base of the solid varies. The region is defined by the area enclosed between $\displaystyle y=x+4$and $\displaystyle y=x^2-2$, therefore, $\displaystyle w=(x+4)-(x^2-2)=6+x-x^2$.

Since the region is bounded by $\displaystyle y=x+4$ and $\displaystyle y=x^2-2$, the base has the following domain: $\displaystyle -2\leq x\leq 3$.

Putting this all together, we find the following:

$\displaystyle V=\int_{-2}^{3}(3)(6+x-x^2) dx$ $\displaystyle = 3[6x+\frac{1}{2}x^2-\frac{1}{3}x^3] |_{-2}^3=3(18+\frac{9}{2}-9+12-2-\frac{8}{3})=62.5$

The cross-sections are perpendicular to the $\displaystyle x$ axis; therefore, the volume expression will be in terms of $\displaystyle x$. 

The area of a square is $\displaystyle A=s^2$, where s is the side length of the square. By applying this formula to our general volume formula $\displaystyle (V=\int_{a}^{b}A(x) \: dx)$, we get the following: $\displaystyle V=\int_{a}^{b}s^2 \: dx$.

Next, an expression for s2 must be determined. Because s should be the width of the solid’s base, the expression for that length can be used to solve for s2. 

$\displaystyle s=(x-3)-(5-x)=2x-8$

$\displaystyle s^2=(2x-8)^2=4x^2-32x+64$

Since the region is bounded by $\displaystyle y=x-3$ and $\displaystyle y=5-x$, the base has the following domain: $\displaystyle 3x5$.

Putting this all together, we find the following:

$\displaystyle V=\int_{3}^{5}4x^2-32x+64 \: dx$

The cross-sections are parallel to the $\displaystyle x$ axis; in other words, they are perpendicular to the $\displaystyle y$ axis. This indicates that the expression should be in terms of $\displaystyle y$.

Because the disk is of radius $\displaystyle R$, the base is defined by the following formula: $\displaystyle x^2+y^2=R^2$. 

The area of a square is $\displaystyle A=s^2$, with s being the side length. By applying this formula to our general volume formula $\displaystyle (V=\int_{a}^{b}A(y) \: dy)$, we get the following: $\displaystyle V=\int_{a}^{b}s^2 \: dy$.

The radius $\displaystyle R$ defines the bounds as being $\displaystyle -R\leq y\leq R$. Next, $\displaystyle s$ can be found by understanding that $\displaystyle s$ differs as the width of the circle changes. The value of $\displaystyle s$ is the distance from one side of the circle to the other at any given point along $\displaystyle -R\leq y\leq R$. The length of one side of the square, therefore, is $\displaystyle s=2\sqrt{R^2-y^2}$.

Putting it all together, the following is obtained:

$\displaystyle V=\int_{-R}^{R}(2\sqrt{R2-y^2})^2\: dy= 4\int_{-R}^{R}R^2-y^2 dy$

The cross-sections are perpendicular to the $\displaystyle x$ axis; therefore, the volume expression will be in terms of $\displaystyle x$. 

The area of a square is $\displaystyle A=s^2$, where s is the side length of the square. By applying this formula to our general volume formula $\displaystyle (V=\int_{a}^{b}A(x) \: dx)$, we get the following: $\displaystyle V=\int_{a}^{b}s^2\: dx$.

Next, an expression for $\displaystyle s^2$ must be determined. Because s should be the width of the solid’s base, the expression for that length can be used to solve for $\displaystyle s^2$. 

$\displaystyle s=e^x-0$

$\displaystyle s^2=e^{2x}$

Since the region is bounded by $\displaystyle x=0$and $\displaystyle x=4$, the base has the following domain: $\displaystyle 0\leq x\leq 4$.

Putting this all together, we find the following:

$\displaystyle V=\int_{0}^{4}e^{2x} \: dx =[\frac{1}{2}e^{2x}] |_0^4=\frac{1}{2}(e^8-e^0)=\frac{e^8-1}{2}$

The cross-sections are perpendicular to the $\displaystyle x$ axis; therefore, the volume expression will be in terms of $\displaystyle x$. 

The area of a rectangle is $\displaystyle A=l*w$. By applying this formula to our general volume formula $\displaystyle (V=\int_{a}^{b}A(x) \: dx)$, we get the following: $\displaystyle V=\int_{a}^{b}lw \: dx$.

Next, an expression for $\displaystyle lw$ must be determined. The problem specifies the length (or height) of the rectangle cross-sections is $\displaystyle 10$. This just leaves the value of $\displaystyle w$ to be found. The width of the rectangle will vary as the region creating the base of the solid varies. The region is defined by the area enclosed between $\displaystyle y=x^2-9$ and $\displaystyle y=0$, therefore, $\displaystyle w=(0)-(x^2-9) =9-x^2$.

Since the region is bounded by $\displaystyle y=x^2-9$ and $\displaystyle y=0$, the base has the following domain: $\displaystyle -3\leq x\leq 3$.

Putting this all together, we find the following:

$\displaystyle V=\int_{-3}^{3}10(9-x^2) dx =10[9x-\frac{1}{3}x^3] |_{-3}^3=10(27-9+27-9)=360$

The cross-sections are perpendicular to the $\displaystyle y$ axis; therefore, the volume expression will be in terms of $\displaystyle y$. 

The area of a rectangle is $\displaystyle A=l*w$. By applying this formula to our general volume formula $\displaystyle (V=\int_{a}^{b}A(y) \: dy)$, we get the following: $\displaystyle V=\int_{a}^{b}lw \: dy$.

Next, an expression for $\displaystyle lw$ must be determined. The problem specifies the length (or height) of the rectangle cross-sections is $\displaystyle 6$. This just leaves the value of $\displaystyle w$ to be found. The width of the rectangle will vary as the region creating the base of the solid varies. The region is defined by the area enclosed between $\displaystyle y=x2+3$, $\displaystyle x=0$ and $\displaystyle y=12$. The function $\displaystyle y=x^2+3$ is rewritten in terms of $\displaystyle y$ to become $\displaystyle x=\sqrt{y-3}$, because the final expression should reflect the fact that the cross sections should be written in terms of $\displaystyle y$. Therefore, $\displaystyle w=\sqrt{y-3} -0 =\sqrt{y-3 }$.

Since the region is bounded by $\displaystyle y=x^2+3$ and $\displaystyle y=12$, the base has the following domain: $\displaystyle 3\leq y\leq 12$.

Putting this all together, we find the following:

$\displaystyle V=\int_{3}^{12}6(\sqrt{y-3}) dy$ $\displaystyle =6\int_{3}^{12}(y-3)^{\frac{1}{2}} dy =6[\frac{2}{3}(y-3)^{\frac{3}{2}}] |_3^{12} =4(27-0)=108$

The cross-sections are perpendicular to the $\displaystyle x$ axis; therefore, the volume expression will be in terms of $\displaystyle x$. 

The area of a rectangle is $\displaystyle A=l*w$. By applying this formula to our general volume formula $\displaystyle (V=\int_{a}^{b}A(x) \: dx)$, we get the following: $\displaystyle V=\int_{a}^b{}lw \: dx$.

Next, an expression for $\displaystyle lw$ must be determined. The problem specifies the length (or height) of the rectangle cross-sections is three times the value of the width, or $\displaystyle l=3w$. The volume expression can now be modified: $\displaystyle V=\int_{a}^{b}3w^2 \: dx$

This just leaves the value of $\displaystyle w$ to be found. The width of the rectangle will vary as the region creating the base of the solid varies. The region is defined by the area enclosed between $\displaystyle y=ln(x)+2$ and $\displaystyle y=0$, along $\displaystyle 1\leq x\leq 4$. Therefore, $\displaystyle w=ln(x)+2-0$.

Putting this all together, we find the following:

$\displaystyle V=\int_{1}^{4}3(ln(x)+2)^2 dx=3\int_{1}^{4}ln(x)^2+4ln(x)+4 \: dx$

The cross-sections are perpendicular to the $\displaystyle x$ axis; therefore, the volume expression will be in terms of $\displaystyle x$. 

The area of a square is $\displaystyle A=s^2$, where $\displaystyle s$ is the side length of the square. By applying this formula to our general volume formula $\displaystyle (V=\int_{a}^{b}A(x) \: dx)$, we get the following: $\displaystyle V=\int_{a}^{b}s^2\: dx$.

Next, an expression for $\displaystyle s^2$ must be determined. Because s should be the width of the solid’s base, the expression for that length can be used to solve for $\displaystyle s^2$. 

$\displaystyle s=(4)-(x^5-x^3+4)=x^3-x^5$

$\displaystyle s^2=(x^3-x^5)^2=x^6-2x^8+x^{10}$

Since the region is bounded by $\displaystyle y=x^5-x^3+4$ and $\displaystyle y=4$, the base has the following domain: $\displaystyle 0\leq x\leq 1$.

Putting this all together, we find the following:

$\displaystyle V=\int_{0}^{1}x^6-2x^8+x^{10} \: dx$

The cross-sections are perpendicular to the $\displaystyle x$ axis; therefore, the volume expression will be in terms of $\displaystyle x$. 

The area of a square is $\displaystyle A=s^2$, where $\displaystyle s$ is the side length of the square. By applying this formula to our general volume formula $\displaystyle (V=\int_{a}^{b}A(x) \: dx)$, we get the following: $\displaystyle V=\int_{a}^{b}s^2 \: dx$.

Next, an expression for $\displaystyle s^2$ must be determined. Because $\displaystyle s$ should be the width of the solid’s base, the expression for that length can be used to solve for $\displaystyle s^2$. 

$\displaystyle s=sin(x)-0$

$\displaystyle s^2=sin^2(x)$

Putting this all together, we find the following:

$\displaystyle V=\int_{\frac{\pi}{2}}^{\pi}sin^2(x) \: dx$