Application of Integrals

Help Questions

AP Calculus BC › Application of Integrals

Questions 1 - 10

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 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

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:

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

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

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:

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

Find the total distance traveled by a particle along the curve from to .

$\int_0^2\sqrt{1+4x^2}dx$

$\int_0^2\sqrt{1+2x^2}dx$

$\int_0^2\sqrt{1+4x^2}dx$

$\int_0^2\sqrt{1+x^2}dx$

$\int_0^2\sqrt{1+2x}dx$

Explanation

To find the required distance, we can use the arc length expression given by $\int_{x_0}^{x_1}\sqrt{1+f'(x)^2}dx$ .

Taking the derivative of our function, we have . Plugging in our values for our integral bounds, we have

$\int_0^2\sqrt{1+(2x)^2}dx = \int_0^2 \sqrt{1+4x^2}dx$ .

As with most arc length integrals, this integral is too difficult (if not, outright impossible) to evaluate explicitly by hand. So we will just leave it this form, or evaluate it with some computer software.

Determine the length of the following function between $1\leq t\leq 4$

$v(t)=\frac{2}{3}(t-1)^{\frac{3}{2}}$

$\frac{14}{3}$

$-\frac{14}{3}$

$\frac{7}{3}$

$-\frac{7}{3}$

$\frac{16}{9}$

Explanation

In order to begin the problem, we must first remember the formula for finding the arc length of a function along any given interval:

$L=\int ds$

where ds is given by the equation below:

$ds=\sqrt{1+(\frac{dy}{dx})^{2}}dx$

We can see from our equation for ds that we must find the derivative of our function, which in our case is dv/dt instead of dy/dx, so we begin by differentiating our function v(t) with respect to t:

$\frac{dv}{dt}=(t-1)^{\frac{1}{2}}=\sqrt{t-1}$

Now we can plug this into the given equation to find ds:

$ds=\sqrt{1+(\frac{dv}{dt})^{2}}dt=\sqrt{1+(\sqrt{t-1})^{2}}=\sqrt{1+t-1}=\sqrt{t}$

Our last step is to plug our value for ds into the equation for arc length, which we can see only involves integrating ds. The interval along which the problem asks for the length of the function gives us our limits of integration, so we simply integrate ds from t=1 to t=4:

$L=\int ds=\int_{1}^{4}\sqrt{t}dt=[\frac{2}{3}t^{\frac{3}{2}}]_{1}^{4}$

$=\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(1)^{\frac{3}{2}}=\frac{16}{3}-\frac{2}{3}=\frac{14}{3}$

Page 1 of 5

Return to subject