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

To determine the volume of a solid with defined cross sectional areas, the equation

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

where $\displaystyle A(x)$ is the cross sectional area at a given x, and the volume exists on the interval $\displaystyle a\leq x \leq b$.  

Because the cross sectional area is a quarter of a circle with a radius of 2x, we find

$\displaystyle A(x)=\frac{1}{4} \pi r^{2}=\frac{1}{4} \pi (2x)^{2}=\pi x^{2}$

The volume is then found with

$\displaystyle V=\int_{0}^{3}\pi x^{2}dx=\frac{1}{3}\pi x^{3}\left.\begin{matrix} \\ \end{matrix}\right|\begin{matrix} \\ \end{matrix}^{3}_{0}=9\pi$

The volume of solid region rotated around the x-axis such as the one in this problem can be found by summing the area of discs, using the formula :

$\displaystyle V=\int_{a}^{b} \pi [f(x)]^{2}dx$

where f(x) gives the radius of each disk.  Applying the equation for this problem:

$\displaystyle V=\int_{0}^{\pi}\pi [(1-\cos ^{2}x)^{1/4}]^{2}dx=\int_{0}^{\pi}\pi (\sin ^{2}x)^{1/2}dx=\int_{0}^{\pi}\pi \sin (x) dx$

$\displaystyle V=-\pi\cos(x)\left.\begin{matrix} \\ \end{matrix}\right|^{\pi}_{0}=2\pi$

This volume can be found with the washer method, which is done using the equation

$\displaystyle V=\int^{b}_{a}\pi[(f(x))^{2}-(g(x))^{2}]dx$

where the region to be revolved around the x axis has the upper bound of f(x), the lower bound of g(x), and exists on the interval [a, b].  Applying this equation:

$\displaystyle V=\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \pi [(\sqrt{2\cos x} )^{2}-(\sqrt{\cos x} )^{2}]dxV=\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \pi \cos(x)dx$

$\displaystyle V=\pi \sin(x)\left.\begin{matrix} \\ \end{matrix}\right|^{\frac{\pi}{2}}_{-\frac{\pi}{2}}=2\pi$

This volume can be found using the shell method by implementing the formula

$\displaystyle V=\int^{a}_{b}2 \pi x [f(x)-g(x)]dx$

Where the region to be rotated about the y-axis is between f(x) and g(x) on the interval $\displaystyle a \leq x \leq b$.  

Additionally, one can notice that here, because f(x)=-g(x), the region between the two function is the same as twice the region betwee f(x) and the x-axis, simplifying our problem.  Using this symmetry and the above formula:

$\displaystyle V=2 \int^{2}_{1}2\pi x (f(x))dx=4\pi \int^{2}_{1}x(x-3)^{2}dx=4\pi \int^{2}_{1}(x^{3}-6x^{2}+9x)dx$

$\displaystyle V=4\pi(\frac{x^{4}}{4}-2x^{3}+\frac{9x^{2}}{2})\left.\begin{matrix} \\ \end{matrix}\right|^{2}_{1}=13\pi$

In order to evaluate an integral by parts, utilize the equation:

$\displaystyle \int u\, dv=uv-\int v \, du$

For the integral given, we can define u and dv:

$\displaystyle u=\frac{1}{2}x^{2}, \,\,\,\, dv=e^{2x}dx$

From this, we can evalute du and v:

$\displaystyle v=\frac{1}{2}e^{2x}, \,\,\,\, du=x\,dx$

Combining these elements usng the equation above, we can begin to evaluate the integral:

$\displaystyle \int \frac{1}{2}x^{2}e^{2x}dx=(\frac{1}{2}x^{2})(\frac{1}{2}e^{2x})-\int \frac{1}{2}e^{2x}x\, dx=\frac{1}{4}x^{2}e^{2x}-\int \frac{1}{2}e^{2x}x\, dx$

The final term of this answer needs to be further, again by integration by parts:

$\displaystyle u=\frac{1}{2}x, \,\,\,\, dv=e^{2x}dx$

$\displaystyle v=\frac{1}{2}e^{2x}, \,\,\,\, du=\frac{1}{2}dx$

Doing the integration:

$\displaystyle \int \frac{1}{2}e^{2x}x\, dx=\frac{1}{4}e^{2x}x-\int\frac{1}{4}e^{2x}dx=\frac{1}{4}e^{2x}x-\frac{1}{8}e^{2x}$

Substituting this in back for the previous partial solution:

$\displaystyle \int \frac{1}{2}e^{2x}x^{2}\, dx=\frac{1}{4}x^{2}e^{2x}-\frac{1}{4}e^{2x}x+\frac{1}{8}e^{2x}+C=\frac{e^{2x}}{4}(x^{2}-x+\frac{1}{2})+C$