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Calculus 2 : Integral Applications

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Example Questions

Example Question #282 : Integrals

Find the area under the given curve, o the interval $0\leq x\leq2$ :

$y(x)=e^{2x}+x^2+2$

Possible Answers:

$\frac{e^4-1}{2}+20/3$

$\frac{e^4}{2}+20/3$

$\frac{e^4}{2}$

Correct answer:

$\frac{e^4-1}{2}+20/3$

Explanation:

The area under the given curve is found using the following integral:

$\int_{0}^{2}(e^{2x}+x^2+2)dx=\frac{e^{2*2}}{2}+\frac{2^3}{3}+2*2-\frac{e^0}{2}=\frac{e^4-1}{2}+20/3$

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Example Question #112 : Integral Applications

Find the area of the region bounded by the curves $\small x = -y^2$ and $\small y = -x-2$

Area under curve problem 11

Possible Answers:

$\frac{13}{6}$

$\small \frac{9}{2}$

$\frac{7}{3}$

Correct answer:

$\small \frac{9}{2}$

Explanation:

Find the area of the region bounded by the curves $\small x = -y^2$ and $\small y = -x-2$

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Definition

The area between two curves over the interval $a\leq y\leq b$ is defined by the integral:

$Area =\int_a^b [f(y)_{upper }-g(y)_{lower}]dy$

where $f(y)_{upper}\geq g(y)_{lower}$ for all such that $a\leq y\leq b$

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Looking at the graphs of the curves we were given, we quickly see that integrating with respect to would be easiest. Attempting to integrate with respect to , we would not be able to assert $f(x)_{upper}\geq g(x)_{lower}$ for all values of between the intersection points. In fact, $\small x = -y^2$ would not even pass the vertical line test to be considered a function.

First let's find where the curves intersect. The coordinate of both points will be our limits of integration. Substitute the equation for the parabola into the equation of the line we were given:

y=(-y^2)-2

y^2 - y-2=0

(y+1)(y-2)=0

The solutions are therefore $y = \left \{ -1,2\right \}$

Area under curve problem 11

$Area =\int_a^b [f(y)_{upper }-g(y)_{lower}]dy$

$\small Area = \int_{-1}^2[-y^2-(-y-2)]dy$

$\small \small Area = \int_{-1}^2[-y^2+y+2)]dy$

$\small \small \small \small Area =\left[-\frac{1}{3}y^3+\frac{1}{2}y^2+2y \right ]_{-1}^{2}$

$\small Area = \left(-\frac{8}{3}+2+4 \right)-\left(\frac{1}{3}+\frac{1}{2}-2 \right )$

$\small \small Area = \left(\frac{10}{3} \right)-\left(-\frac{7}{6} \right )$

$\small Area = \frac{9}{2}$

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Example Question #1 : Volume Of A Solid

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

y=x^2-4x+5

y=0

x=1

x=4

Possible Answers:

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

$18\pi$

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

$32\pi$

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

Correct answer:

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

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:

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

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

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Example Question #1 : Volume Of Cross Sections And Area Of Region

Suppose the functions f(x) = 0 , x=y^2 , and x=4 form a closed region. Rotate this region across the x-axis. What is the volume?

Possible Answers:

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

$8\pi$

$4\pi$

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

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

Correct answer:

$8\pi$

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.

r=y

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

h=4-x=4-y^2

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

4=y^2

2=y

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

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

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

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

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Example Question #3 : Volume Of Cross Sections And Area Of Region

Find the volume of the solid generated by revolving the region bounded by {}y = sin(x) and the -axis in the first quadrant about the -axis.

Possible Answers:

$\frac{1}{2}\pi^2$

$\frac{1}{2}\pi$

$\frac{1}{4}\pi$

$\pi^2$

$\pi$

Correct answer:

$\frac{1}{2}\pi^2$

Explanation:

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

Recall the formula:

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

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

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

Using trigonometric identities, we know that:

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

Hence:

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

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

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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 #1 : 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:

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

$V=215\pi$ units cubed

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

$V=\frac{512\pi}{3}$ 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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Calculus 2 : Integral Applications

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #282 : Integrals

Example Question #112 : Integral Applications

Example Question #1 : Volume Of A Solid

Example Question #1 : Volume Of Cross Sections And Area Of Region

Example Question #3 : Volume Of Cross Sections And Area Of Region

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 #1 : Volume Of Cross Sections And Area Of Region

All Calculus 2 Resources

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