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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Volume Of A Solid

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 ${}\small t=0$ .

Possible Answers:

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

$\small 27$

$\small 270$

$\small 90$

$\small 15$

Correct answer:

$\small 90$

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.

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Example Question #1 : Volume Of A Solid

Approximate the volume of a solid in the first quadrant revolved about the y-axis and bounded by the functions: y=9x and y=x^3 . Round the volume to the nearest integer.

Possible Answers:

Correct answer:

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.

9x=x^3

9=x^2

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.

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

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

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:

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Example Question #2 : Volume Of A Solid

Determine the volume of a solid created by rotating the curve $y=\sqrt[3]{x}$ and the line x=1 by revolving around the -axis.

Possible Answers:

$\frac{8}{5}\pi$

$\frac{12}{5}\pi$

$\frac{3}{5}\pi$

$\frac{2}{5}\pi$

$\frac{1}{5}\pi$

Correct answer:

$\frac{3}{5}\pi$

Explanation:

Write the volume formula for cylindrical shells.

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

The shell radius is .

The shell height is the function in terms of . Rewrite that equation.

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

x=y^3

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

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

Substitute all the values and solve for the volume.

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

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

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Example Question #1 : Volume Of A Solid

What is the volume of the solid formed when the line $y=\sqrt{x}$ is rotated around the -axis from x=4 to x = 9 ?

Possible Answers:

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

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

$36\pi$

Correct answer:

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

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 x=y^2 .

We'll also need to convert the endpoints of the interval to y-values. Note that when $x=4, \ y=2$ , and when x = 9, y = 3. 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 f(x) 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

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

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

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

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

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Example Question #6 : Volume Of Cross Sections And Area Of Region

Find the volume of the solid generated when the function

f(x)=x

is revolved around the x-axis on the interval [0,8] .

Hint: Use the method of cylindrical disks.

Possible Answers:

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

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

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

$V=215\pi$ units cubed

Correct answer:

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

Explanation:

The formula for the volume is given as

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

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

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

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Example Question #1 : Volume Of A Solid

Find the volume V of a solid whose cross section at x is a quarter circle with radius 2x on the interval [0, 3].

Possible Answers:

$V=\frac{9}{2}\pi$

$V=9\pi$

$V=18\pi$

$V=\frac{9}{4}\pi$

Correct answer:

$V=9\pi$

Explanation:

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

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

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

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

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

The volume is then found with

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

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Example Question #10 : Volume Of A Solid

Let R be the region between the graph of $f(x)=(1-\cos^{2}x)^{1/4}$ and the x-axis on the interval $0\leq x \leq \pi$ . Find the volume V of the solid obtained by revolving R about the x-axis.

Possible Answers:

$V=\frac{\pi}{2}$

$V=2\pi$

V=0

$V=\pi$

$V=4\pi$

Correct answer:

$V=2\pi$

Explanation:

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 :

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

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

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

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

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Example Question #11 : Volume Of A Solid

Let R be the region between f(x) and g(x) on the given interval. FInd the volume V obtained by revolving R around the x-axis.

$f(x)=\sqrt{2 \cos(x)}$

$g(x)=\sqrt{ \cos(x)}$

Interval: $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$

Possible Answers:

$V=\pi$

$V=\frac{\pi}{2}$

$V=4\pi$

$V=\frac{\pi}{4}$

$V=2\pi$

Correct answer:

$V=2\pi$

Explanation:

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

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

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:

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

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

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Example Question #12 : Volume Of A Solid

Let R be the region between the function f(x) and g(x) on the interval given below. Find the volume V obtained by revolving R around the y-axis.

$f(x)=(x-3)^{2}$

$g(x)=-(x-3)^{2}$

Interval: $1 \leq x \leq 2$

Possible Answers:

$V=\frac{14\pi}{3}$

$V=13\pi$

$V=\frac{28\pi}{3}$

$V=\frac{13\pi}{2}$

$V=\frac{26\pi}{3}$

Correct answer:

$V=13\pi$

Explanation:

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

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

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

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:

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

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

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Example Question #13 : Volume Of A Solid

Using integration by parts, evaluate the following indefinite integral:

$\int \frac{1}{2}x^{2}e^{2x}dx$

Possible Answers:

$\frac{e^{2x}}{2}(x^{2}-x+\frac{1}{2})+C$

$\frac{e^{2x}}{4}(x-\frac{1}{2})+C$

$\frac{1}{4}x^{2}e^{2x}+C$

$\frac{e^{2x}}{4}(x^{2}-x+\frac{1}{2})+C$

$\frac{e^{2x}}{2}(x^{2}+x-\frac{1}{2})+C$

Correct answer:

$\frac{e^{2x}}{4}(x^{2}-x+\frac{1}{2})+C$

Explanation:

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

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

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

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

From this, we can evalute du and v:

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

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

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

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

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

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

Doing the integration:

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

Substituting this in back for the previous partial solution:

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

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Volume Of A Solid

Example Question #1 : Volume Of A Solid

Example Question #2 : Volume Of A Solid

Example Question #1 : Volume Of A Solid

Example Question #6 : Volume Of Cross Sections And Area Of Region

Example Question #1 : Volume Of A Solid

Example Question #10 : Volume Of A Solid

Example Question #11 : Volume Of A Solid

Example Question #12 : Volume Of A Solid

Example Question #13 : Volume Of A Solid

All Calculus 2 Resources

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