From calculus, we know the volume of an irregular solid can be determined by evaluating the following integral:

$\displaystyle V=\int_{a}^{b}A(x)dx$

Where A(x) is an equation for the cross-sectional area of the solid at any point x. We know our bounds for the integral are x=1 and x=4, as given in the problem, so now all we need is to find the expression A(x) for the area of our solid.

From the given bounds, we know our unrotated region is bounded by the x-axis (y=0) at the bottom, and by the line y=x^2-4x+5 at the top. Because we are rotating about the x-axis, we know that the radius of our solid at any point x is just the distance y=x^2-4x+5. Now that we have a function that describes the radius of the solid at any point x, we can plug the function into the formula for the area of a circle to give us an expression for the cross-sectional area of our solid at any point:

$\displaystyle A(x)=\pi r^2=\pi (x^2-4x+5)^2=\pi (x^4-8x^3+26x^2-40x+25)$

We now have our equation for the cross-sectional area of the solid, which we can integrate from x=1 to x=4 to find its volume:

$\displaystyle V=\int_{1}^{4}\pi (x^4-8x^3+26x^2-40x+25)dx$

$\displaystyle =\pi [\frac{1}{5}x^5-2x^4+\frac{26}{3}x^3-20x^2+25x]_{1}^{4}=\frac{78\pi }{5}$

Write the formula for cylindrical shells, where $\displaystyle r$ is the shell radius and $\displaystyle h$ is the shell height.

$\displaystyle V=\int_{a}^{b} 2\pi rh \:dy$

Determine the shell radius.

$\displaystyle r=y$

Determine the shell height.  This is done by subtracting the right curve, $\displaystyle x=4$, with the left curve, $\displaystyle x=y^2$.

$\displaystyle h=4-x=4-y^2$

Find the intersection of $\displaystyle x=y^2$ and $\displaystyle x=4$ to determine the y-bounds of the integral.

$\displaystyle 4=y^2$

$\displaystyle 2=y$

The bounds will be from 0 to 2.  Substitute all the givens into the formula and evaluate the integral.

$\displaystyle V=\int_{a}^{b} 2\pi rh \:dy = \int_{0}^{2} 2\pi \cdot y\cdot \left ( 4-y^2)dy$

$\displaystyle V= 2\pi \int_{0}^{2} \left ( 4y-y^3\right ) dy$

$\displaystyle V=2\pi \left[ 2y^2 - \frac{1}{4}y^4\right]_{0}^{2} = 2\pi \left[2(2)^2 -\frac{1}{4}(2)^4\right] = 2\pi[8-4]= 8\pi$

Since we are revolving a region of interest around a horizontal line $\displaystyle {}y = 0$, we need to express the inner and outer radii in terms of x.

Recall the formula:

$\displaystyle Volume = \pi \int_{b}^{a} [(\textup{outer radius})^{2} - (\textup{inner radius})^{2}]dx$

The outer radius is $\displaystyle y = sin(x)$ and the inner radius is $\displaystyle y = 0$. The x-limits of the region are between $\displaystyle x=0$ and $\displaystyle x=\pi$. So the volume set-up is:

$\displaystyle V = \pi \int_{0}^{\pi} sin^{2}x$

Using trigonometric identities, we know that:    

$\displaystyle sin^{2}x = \frac{1}{2} - \frac{1}{2}cos 2x$

Hence:

$\displaystyle V = \pi \int_{0}^{\pi} sin^2x dx =\pi \int_{0}^{\pi} \left(\frac{1}{2} - \frac{1}{2}cos2x\right) dx$

$\displaystyle = \pi \left[\frac{1}{2}x - \frac{1}{4}sin2x\right]_{0}^{\pi} = \frac{1}{2}\pi^2$

The rate at which the height changes is $\displaystyle \small \frac{\mathrm{d} h}{\mathrm{d} t}$, which means $\displaystyle \small \small \frac{\mathrm{d} h}{\mathrm{d} t}=5t^{1/2}$.

To find the height after nine seconds, we need to integrate to get $\displaystyle {}\small h(t)$.

We can multiply both sides by $\displaystyle \small dt$ to get $\displaystyle \small dh=5t^{1/2}dt$ and then integrate both sides.

This gives us 

$\displaystyle \small h=\frac{10t^{3/2}}{3}+c$.

Since the cup is empty at $\displaystyle \small t=0, h(0)=0$, so $\displaystyle \small c=0$.

This means $\displaystyle \small \small h(9)=10(9)^{3/2}/3=90$. No units were given in the problem, so leaving the answer unitless is acceptable.

Write the washer's method.

$\displaystyle V=\int_{a}^{b}\pi[R(y)^2-r(y)^2]\:dy$

Set the equations equal to each other to determine the bounds.

$\displaystyle 9x=x^3$

$\displaystyle 9=x^2$

$\displaystyle x=3$

The bounds are from 0 to 3.

Determine the big and small radius.  Rewrite the equations so that they are in terms of y.

$\displaystyle R(y)= \sqrt[3]{y}$

$\displaystyle r(y)=\frac{y}{9}$

Set up the integral and solve for the volume.

$\displaystyle V=\int_{0}^{3}\pi[(\sqrt[3]{y})^2-(\frac{y}{9})^2]\:dy$

$\displaystyle V=\int_{0}^{3}\pi[y^{\frac{2}{3}}-\frac{y^2}{81}]\:dy=\pi[\frac{3}{5}y^\frac{5}{3}-\frac{y^3}{243}]_{0}^{3}=11.41353$

The volume to the nearest integer is:  $\displaystyle 11$

Write the volume formula for cylindrical shells.

$\displaystyle V=\int_{a}^{b}2\pi (\textup{shell radius})(\textup{shell height}) dy$

The shell radius is $\displaystyle y$.

The shell height is the function in terms of $\displaystyle y$.  Rewrite that equation.

$\displaystyle y=\sqrt[3]{x}$

$\displaystyle x=y^3$

$\displaystyle \textup{shell height}: 1-y^3$

The bounds lie on the y-axis since the thickness variable is $\displaystyle dy$.  This is from 0 to 1, since the intersection of the line $\displaystyle x=1$ and $\displaystyle y=\sqrt[3]{x}$ is at $\displaystyle (1,1)$.  

Substitute all the values and solve for the volume.

$\displaystyle V=\int_{0}^{1}2\pi (y)(1-y^3) dy=\int_{0}^{1}2\pi (y-y^4)dy$

$\displaystyle V=2\pi \int_{0}^{1}(y-y^4)dy= 2\pi \left[\frac{y^2}{2}-\frac{y^5}{5}\right]_{0}^{1}= 2\pi\left[\frac{1}{2}-\frac{1}{5}\right]= \frac{3}{5}\pi$

To rotate a curve around the y-axis, first convert the function so that y is the independent variable by solving $\displaystyle y=\sqrt{x}$ for x. This leads to the function $\displaystyle x=y^2$. 

We'll also need to convert the endpoints of the interval to y-values. Note that when $\displaystyle x=4, \ y=2$, and when $\displaystyle x = 9, y = 3.$ Therefore, the the interval being rotated is from $\displaystyle y=2\ to\ y=3$.

The disk method is best in this case. The general formula for the disk method is 

$\displaystyle V=\pi\int_{a}^{b}f^2(x)dx$, where V is volume, $\displaystyle a\ and\ b$ are the endpoints of the interval, and $\displaystyle f(x)$ the function being rotated. 

Substuting the function and endpoints from the problem at hand leads to the integral

$\displaystyle V=\pi\int_{2}^{3}(y^2)^2dy=\pi\int_{2}^{3}y^4dy$.

To evaluate this integral, you must know the power rule. Recall that the power rule is 

$\displaystyle \int{y^ndy}=\frac{y^{n+1}}{n+1}+c$. 

$\displaystyle \pi\int_{2}^{3}y^4dy=\pi[\frac{y^5}{5}]_{2}^{3}$

$\displaystyle =\pi\left[\frac{3^5}{5}-\frac{2^5}{5}\right]=\pi\left[\frac{243}{5}-\frac{32}{5}\right]$

$\displaystyle =\frac{211\pi}{5}$.

The formula for the volume is given as

$\displaystyle V=\pi \int_{a}^{b} r^2 \, dx$

where $\displaystyle r=f(x)$ and the bounds on the integral come from the interval $\displaystyle [a,b]$.

As such, 

$\displaystyle V=\pi \int_{0}^{8} x^2 \, dx$

When taking the integral, we will use the inverse power rule which states

$\displaystyle \int x^n=\frac{x^{n+1}}{n+1}$

Applying this rule we get

$\displaystyle \pi \int_{0}^{8} x^2 \, dx=\pi \left(\frac{x^3}{3}\right) \left.\begin{matrix} \\ \end{matrix}\right|_{0}^{8}$

And by the corollary of the First Fundamental Theorem of Calculus

$\displaystyle \pi \left(\frac{x^3}{3}\right) \left.\begin{matrix} \\ \end{matrix}\right|_{0}^{8}=\pi \left(\left[\frac{8^3}{3}\right]-\left[\frac{0^3}{3}\right]\right)$

$\displaystyle =\frac{512\pi}{3}$

As such,

$\displaystyle V=\frac{512\pi}{3}$ units cubed

The formula for volume of the region revolved around the x-axis is given as 

$\displaystyle V=\pi \int_{a}^{b} r^2\, dx$

where $\displaystyle r=f(x)$

As such

$\displaystyle V=\pi \int_{0}^{10} (-x^2+10x+3)^2\,dx = \pi \int_{0}^{10}x^4-20x^3+94x^2+60x+9\,dx$

When taking the integral, we use the inverse power rule which states

$\displaystyle \int x^n\,dx = \frac{x^{n+1}}{n+1}$

Applying this rule term by term we get

$\displaystyle = \pi \int_{0}^{10}x^4-20x^3+94x^2+60x+9\,dx = \pi\left ( \frac{x^5}{5} - \frac{20x^4}{4} + \frac{94x^3}{3} + \frac{60x^2}{2} + 9x\right |_{0}^{10}$

And by the corollary of the Fundamental Theorem of Calculus 

$\displaystyle \pi\left ( \frac{x^5}{5} - \frac{20x^4}{4} + \frac{94x^3}{3} + \frac{60x^2}{2} + 9x\right |_{0}^{10}=\pi \left ( \left [ \frac{(10)^5}{5}-\frac{20(10)^4}{4}+\frac{94(10)^3}{3}+\frac{60(10)^2}{2}+9(10) \right ]-0 \right )$

$\displaystyle =\frac{13270}{3}\pi$

As such the volume is

$\displaystyle \frac{13270}{3}\pi$ units cubed

Since we are revolving a function of $\displaystyle x$ around the y-axis, we will use the method of cylindrical shells to find the volume.

Using the formula for cylindrical shells, we have

$\displaystyle \int_{0}^{4}2\pi x (\sqrt{x})dx$

$\displaystyle =\int_{0}^{4}2\pi x^{3/2}dx$

$\displaystyle =[2\pi x^{5/2}]^{4}_{0}$

$\displaystyle =\frac{128}{5}\pi$.