AP Calculus BC - Volume of Cross Sections and Area of Region

Find the volume of the solid generated by rotating about the y-axis the region under the curve $y =\sqrt{x}$ , from to .

$\frac{128}{5}\pi$

$\frac{64}{3}\pi$

$\frac{\pi}{2}$

$\frac{256}{7}\pi^2$

None of the other answers

Explanation

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

Using the formula for cylindrical shells, we have

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

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

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

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

Find the volume of the solid generated when the function

is revolved around the x-axis on the interval .

Hint: Use the method of cylindrical disks.

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

$V=\frac{625\pi}{3}$ units cubed

$V=\frac{400\pi}{3}$ units cubed

$V=215\pi$ units cubed

Explanation

The formula for the volume is given as

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

where and the bounds on the integral come from the interval .

As such,

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

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

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

Applying this rule we get

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

And by the corollary of the First Fundamental Theorem of Calculus

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

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

As such,

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

Find the area bound by the curve of g(t), the x and y axes, and the line $t=\frac{3 \pi}{4}$

Explanation

Find the area bound by the curve of g(t), the x and y axes, and the line $t=\frac{3 \pi}{4}$

We are asked to find the area under a curve. This sounds like a job for an integral.

We need to integrate our function from 0 to $t=\frac{3 \pi}{4}$ . These will be our limits of integration. With that in mind, our integral look like this.

$G(t)=\int_0^\frac{3\pi}{4}4sin(t)-6cos(t)dt$

Now, we need to recall the rules for integrating sine and cosine.

$1) \int sin(t)dt=-cos(t)+c$

$2) \int cos(t)dt=sin(t)+c$

With these rules in mind, we can integrate our original function to get:

$G(t)=\int_0^\frac{3\pi}{4}4sin(t)-6cos(t)dt=-4cos(t)-6sin(t)+c|_0^{\frac{3 \pi}{4}}$

$G(t)=-4cos(t)-6sin(t)+c|_0^{\frac{3 \pi}{4}}$

Now, we just need to evaluate our new function at the given limits. Let's start with 0

And our upper limit...

$G({\frac{3 \pi}{4}})=-4cos({\frac{3 \pi}{4}})-6sin({\frac{3 \pi}{4}})+c$

$-4cos({\frac{3 \pi}{4}})-6sin({\frac{3 \pi}{4}})+c=(-4)\frac{-\sqrt{2}}{2}-(6)\frac{-\sqrt{2}}{2}+c$

$(-4)\frac{-\sqrt{2}}{2}-(6)\frac{\sqrt{2}}{2}+c=2\sqrt{2}-3\sqrt{2}+c=-\sqrt{2}+c$

Now, we need to take the difference between our limits:

$G(\frac{3 \pi}{4})-G(0)=(-\sqrt{2}+c)-(-4+c)=-\sqrt{2}+4\approx2.585$

So, our answer is 2.59

Approximate the volume of a solid in the first quadrant revolved about the y-axis and bounded by the functions: and . Round the volume to the nearest integer.

Explanation

Write the washer's method.

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

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

The bounds are from 0 to 3.

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

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

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

Set up the integral and solve for the volume.

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

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

The volume to the nearest integer is:

Suppose the functions , , and form a closed region. Rotate this region across the x-axis. What is the volume?

$8\pi$

$\frac{32\pi}{5}$

$\frac{64}{5}\pi$

$\frac{128\pi}{13}$

$4\pi$

Explanation

Write the formula for cylindrical shells, where is the shell radius and is the shell height.

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

Determine the shell radius.

Determine the shell height. This is done by subtracting the right curve, , with the left curve, .

Find the intersection of and to determine the y-bounds of the integral.

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

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

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

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

What is the volume of the solid formed when the line $y=\sqrt{x}$ is rotated around the -axis from to ?

$\frac{211\pi}{5}$

$\frac{21\pi}{2}$

$36\pi$

Explanation

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

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

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

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

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

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

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

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

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

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

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

A man fills up a cup of water by leaving it outside during a rainstorm. The rate at which the height of the cup changes is equal to $\small 5t^{1/2}$ . What is the height of water at $\small t=9$ ? Assume the cup is empty at ${}\small t=0$ .

$\small 90$

$\small 15$

$\small \frac{5}{6}$

$\small 270$

$\small 27$

Explanation

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

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

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

This gives us

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

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

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

Determine the volume of the solid obtained by rotating the region with the following bounds about the x-axis:

$\frac{78\pi }{5}$

$\frac{36\pi }{7}$

$18\pi$

$32\pi$

$\frac{46\pi }{3}$

Explanation

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

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

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:

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

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:

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

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

Using the method of cylindrical disks, find the volume of the region revolved around the x-axis of the graph of

on the interval

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

$20000\pi$ units cubed

$15000\pi$ units cubed

$10000\pi$ units cubed

Explanation

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

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

where

As such

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

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

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

Applying this rule term by term we get

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

And by the corollary of the Fundamental Theorem of Calculus

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