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Calculus 1 : Regions

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Example Question #71 : Volume

Find the volume of the solid obtained by rotating the region bound by y=x and $y=x^{2}$ about the y-axis using the Shell Method.

Possible Answers:

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

$V=\pi$

$V=\frac{\pi}{6}$

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

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

Correct answer:

$V=\frac{\pi}{6}$

Explanation:

The radius of a single shell can be expressed as , the circumference can be expressed as $2\pi x$ , and the height can be expressed as $x-x^{2}$ .

To obtain the volume of all the shells, we integrate the overall volume. Our bounds can be found by calculating the endpoints, or where $x=x^{2}$ . Thus, our endpoints are at x=0 and x=1 .

Integrating, we get:

$V= \int_0^1 (2\pi x)(x-x^{2})dx=2\pi\int_0^1(x^{2}-x^{3})dx =2\pi\left [ \frac{x^{3}}{3}-\frac{x^4}{4}\right ]_0^1=\frac{\pi}{6}$

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Example Question #72 : Volume

Use the Shell Method to find the volume of the solid obtained by rotating the region under the curve $y=\sqrt{x}$ from to about the x-axis

Possible Answers:

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

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

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

$V=\pi$

$V=\frac{\pi}{6}$

Correct answer:

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

Explanation:

To use shells we relabel the curve $y=\sqrt{x}$ as $x=y^{2}$ .

The radius of a single shell can be expressed as , the circumference can be expressed as $2\pi y$ , and the height can be expressed as $1-y^{2}$ .

To obtain the volume of all the shells, we integrate the overall volume. Our bounds are given as and .

Integrating, we get:

$V=\int_0^1 (2\pi y)(1-y^{2})dy = 2\pi\int_0^1 (y-y^{3})dy =2\pi\left [ \frac{y^{2}}{2}-\frac{y^{4}}{4}\right]_0^1=\frac{\pi}{2}$

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Example Question #72 : Volume

Use the Shell Method to find the volume of the solid obtained by rotating the region under the curve $y=\sqrt{x}$ from to about the x-axis

Possible Answers:

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

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

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

$V=\pi$

$V=\frac{\pi}{6}$

Correct answer:

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

Explanation:

To use shells we relabel the curve $y=\sqrt{x}$ as $x=y^{2}$ .

The radius of a single shell can be expressed as , the circumference can be expressed as $2\pi y$ , and the height can be expressed as $1-y^{2}$ .

To obtain the volume of all the shells, we integrate the overall volume. Our bounds are given as and .

Integrating, we get:

$V=\int_0^1 (2\pi y)(1-y^{2})dy = 2\pi\int_0^1 (y-y^{3})dy =2\pi\left [ \frac{y^{2}}{2}-\frac{y^{4}}{4}\right]_0^1=\frac{\pi}{2}$

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Example Question #73 : Volume

Using the method of cross sections, find the volume of the solid bounded by the function

f(x)=x

x=0

x=9

and the x-axis.

The cross sections are squares perpendicular to the x-axis.

Possible Answers:

243

$\frac{81}{2}$

703

352

Correct answer:

243

Explanation:

The formula for the volume of a solid using the method of cross sections perpendicular to the x-axis on the interval is

Where is the area of the cross section.

Because the cross sections are squares with a side length of the formula becomes

In order to solve the integral we use the inverse power rule which says

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

Applying this rule we get

And by the corollary of the Fundamental Theorem of Calculus we get

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Example Question #71 : How To Find Volume Of A Region

Determine the volume of the solid obtained by rotating the region bounded by x=(y-2)^2 and y=x about the line y=-1 .

Possible Answers:

$\frac{189\pi}{4}$

$\frac{63\pi}{2}$

$\frac{72\pi}{5}$

$\frac{36\pi}{5}$

$32\pi$

Correct answer:

$\frac{63\pi}{2}$

Explanation:

To find the endpoints, we set:

y=(y-2)^2

y=y^2-4y+4

0=y^2-5y+4

0=(y-1)(y-4)

and thus, we get y=4 and y=1 as our bounds.

The area can be found by:

$A(y)=2\pi(radius)(height)=2\pi(y+1)(y-(y-2)^2)=2\pi(-y^3+4y^2+y-4)$

Integrating over our volume, we get:

$V=2\pi\int_1^4(-y^3+4y^2+y-4)dy=2\pi\left [ (-\frac{y^4}{4}+\frac{4y^3}{3}+\frac{y^2}{2}-4y) \right ]_1^4=\frac{63\pi}{2}$

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Example Question #75 : Volume

Determine the volume of the solid obtaining by rotating the region bounded by $y=2\sqrt{x-1}$ and y=x-1 about the line x=6 .

Possible Answers:

$\frac{272\pi}{15}$

$\frac{136\pi}{15}$

$\frac{272\pi}{5}$

$\frac{136\pi}{5}$

$20\pi$

Correct answer:

$\frac{272\pi}{15}$

Explanation:

To find the bounds, we set $(x-1)=2\sqrt{x-1}$ . Solving, we get (x-1)(x-5)=0 , giving us bounds of x=1 and x=5 .

The cross sectional area is then: $A(x)=2\pi(radius)(height)= 2\pi(6-x)(2\sqrt{x-1}-x+1)=2\pi(x^2-7x+6+12\sqrt{x-1}-2x\sqrt{x-1})$

Integrating, we get the volume:

$V=\int_a^b A(x)dx=2\pi\int_1^5 (x^2-7x+6+12\sqrt{x-1}-2x\sqrt{x-1})dx=2\pi\left [ (\frac{x^3}{3}-\frac{7x^2}{2}+6x+8(x-1)^{\frac{3}{2}}-\frac{4}{5}(x-1)^\frac{5}{2}-\frac{4}{3}(x-1)^{\frac{3}{2}})) \right ]_1^5=2\pi(\frac{136}{15})=\frac{272\pi}{15}$

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Example Question #2951 : Functions

Find the volume of the solid fo revolution formed by rotating the region formed by the x-axis and the graph of $y=\sqrt{x}$ from x=0 to x=1, about the y-axis.

Possible Answers:

$\pi$

$\frac{5\pi}{4}$

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

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

$\frac{4\pi}{5}$

Correct answer:

$\frac{4\pi}{5}$

Explanation:

$V=\int_0^1(2\pi x \sqrt{x})dx=2\pi\int_0^1 x^\frac{3}{2}dx=2\pi\left [ \frac{2 x^{\frac{5}{2}}}{5}\right ]_0^1=\frac{4\pi}{5}$

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Example Question #77 : Volume

The region enclosed by the curves y=x and y=x^2 is rotated about the x-axis. Using discs or washers, find the volume of the resulting solid.

Possible Answers:

$\frac{2\pi}{15}$

$\frac{\pi}{3}$

$\frac{\pi}{15}$

Correct answer:

$\frac{2\pi}{15}$

Explanation:

The intersection points are (0,0) and (1,1). The cross-sectional area is calculated by finding the area between curves, using integration, getting: $A(x)=\pi x^2- \pi(x^2)^2=\pi(x^2-x^4)$

The volume is therefore:

$V=\int_0^1 \pi(x^2-x^4)=\left [\pi( \frac{x^3}{3}-\frac{x^5}{5}) \right ]_0^1=\frac{2\pi}{15}$

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Example Question #78 : Volume

The region enclosed by the curves y=x and y=x^2 is rotated about the line y=2 . Using discs and washers, find the volume of the resulting solid.

Possible Answers:

$\frac{\pi}{2}$

$\frac{8\pi}{15}$

$\frac{2\pi}{15}$

$\frac{3\pi}{2}$

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

Correct answer:

$\frac{8\pi}{15}$

Explanation:

The intersection points are (0,0) and (1,1). The cross-sectional area is: $A(x)=\pi(2-x^2)^2-\pi(2-x)^2$

The volume is therefore:

$V=\int_0^1 \pi((2-x^2)^2-(2-x)^2)dx=\pi\int_0^1 (x^4-5x^2+4x)dx$

$=\pi\left [ \frac{x^5}{5}-\frac{5x^3}{3}+2x^2\right ]_0^1=\frac{8\pi}{15}$

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Example Question #79 : Volume

Finding area under a curve is extremely similar to finding the volume. The volume of a cylinder is $V=\pi r^2 h$ . When finding the volume of a function rotated around the x-axis, we will look at summing infinitesimal cylinders (disks). The height of each of these cylinders is , the radius of the cylinder is the function since given any x value, f(x) is the distance from the x-axis to the curve. Thus if we want to find the volume of a function f(x) between [a,b] that is rotated about the x-axis we simply use the equation $\int_a^b \pi f(x)^2 dx$ .

Find the volume of the solid obtained from rotating the function x^3+7 about the x-axis and bounded by the y-axis and x=4 .

Possible Answers:

$\frac{24028\pi}{7}$

$92\pi$

$\frac{24028}{7}$

$\frac{17756\pi}{7}$

Correct answer:

$\frac{24028\pi}{7}$

Explanation:

The bounds are [0,4] since it is bounded by the y-axis and the line x=4 and the function is f(x)=x^3+7 .

Therefore the equation is

$\\ \int_0^4\pi (x^3+7)^2 dx\\ \\= \pi \int_0^4 (x^6+14x^3+49)dx \\ \\= \pi(\frac{1}{7}x^7+\frac{7}{2}x^4+49x)|_0^4$ .

This gives us the answer of

$\frac{24028\pi}{7}$ .

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Calculus 1 : Regions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #71 : Volume

Example Question #72 : Volume

Example Question #72 : Volume

Example Question #73 : Volume

Example Question #71 : How To Find Volume Of A Region

Example Question #75 : Volume

Example Question #2951 : Functions

Example Question #77 : Volume

Example Question #78 : Volume

Example Question #79 : Volume

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