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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #61 : Volume

Using the Shell Method, find the volume of the solid obtained by rotating the region bound by $y=x-x^{2}$ and y=0 about the line x=2

Possible Answers:

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

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

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

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

$V=\pi$

Correct answer:

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

Explanation:

The parabola's x-intercepts are 0 and 1. The radius is 2-x , the circumference is $2\pi(2-x)$ and the height is $(x-x^{2})$ .

Integrating, we get:

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

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Example Question #3981 : Calculus

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:

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

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

$V=\pi$

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

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

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 #3981 : Calculus

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:

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

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

$V=\pi$

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

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

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 #71 : How To Find Volume Of A Region

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:

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

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

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

$V=\pi$

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

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 #71 : How To Find Volume Of A Region

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:

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

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

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

$V=\pi$

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

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 #3982 : Calculus

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:

352

243

703

$\frac{81}{2}$

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 : Volume

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:

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

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

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

$32\pi$

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

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 #71 : 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:

$20\pi$

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

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

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

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

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 #71 : Regions

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:

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

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

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

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

$\pi$

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 #74 : 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:

$\frac{\pi}{15}$

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

$\frac{\pi}{3}$

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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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #61 : Volume

Example Question #3981 : Calculus

Example Question #3981 : Calculus

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

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

Example Question #3982 : Calculus

Example Question #71 : Volume

Example Question #71 : Volume

Example Question #71 : Regions

Example Question #74 : Volume

All Calculus 1 Resources

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