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Calculus 1 : Volume

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #2944 : Functions

Sphere segment

A sphere with a radius of 13.5 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 1.8 units to the left of the origin and the second cut is 3.2 units to the right of the origin, what is the volume of the segment?

Possible Answers:

1009.4

773.4

2822.4

2834.6

352.2

Correct answer:

2822.4

Explanation:

To approach this problem, begin by picturing how the sphere might be projected onto the two-dimensional Cartesian plane:

Sphere segment outlined

In its two-dimensional projection, we notice the outline of a circle. The formula for a circle of radius in the Cartesian coordinate system is given as:

x^2+y^2=r^2

Which can be rewritten in terms of as $y=\sqrt{r^2-x^2}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

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

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{a}^{b}\pi(r^2-x^2)dx$

$V=\pi(r^2x-\frac{x^3}{3})|_{a}^{b}$

Consider the points where circle contacts the x-axis, the greatest and smallest possible values of on the circle:

$V=\pi(r^2x-\frac{x^3}{3})|_{-r}^{r}$

$V=\pi(r^2r-\frac{r^3}{3})-\pi(r^2(-r)-\frac{(-r)^3}{3})$

$V=\pi(\frac{2}{3}r^3)-\pi(-\frac{2}{3}r^3)$

$V=\frac{4}{3}\pi r^3$

This is how the volume of a sphere can be derived, and similary, this is how we can derive the area of our segment:

$V=\pi(13.5^2x-\frac{x^3}{3})|_{-1.8}^{3.2}$

$V=\pi(13.5^2((3.2)-(-1.8))-\frac{(3.2)^3-(-1.8)^3}{3})=2822.4$

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Example Question #61 : How To Find Volume Of A Region

Sphere segment

A sphere with a radius of 25 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 23 units to the left of the origin and the second cut is 3 units to the left of the origin, what is the volume of the segment?

Possible Answers:

$\frac{23582}{3}\pi$

$\frac{11791}{3}\pi$

$\frac{25360}{3}\pi$

$\frac{21214}{3}\pi$

$\frac{12680}{3}\pi$

Correct answer:

$\frac{25360}{3}\pi$

Explanation:

To approach this problem, begin by picturing how the sphere might be projected onto the two-dimensional Cartesian plane:

Sphere segment outlined

In its two-dimensional projection, we notice the outline of a circle. The formula for a circle of radius in the Cartesian coordinate system is given as:

x^2+y^2=r^2

Which can be rewritten in terms of as $y=\sqrt{r^2-x^2}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

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

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{a}^{b}\pi(r^2-x^2)dx$

$V=\pi(r^2x-\frac{x^3}{3})|_{a}^{b}$

Consider the points where circle contacts the x-axis, the greatest and smallest possible values of on the circle:

$V=\pi(r^2x-\frac{x^3}{3})|_{-r}^{r}$

$V=\pi(r^2r-\frac{r^3}{3})-\pi(r^2(-r)-\frac{(-r)^3}{3})$

$V=\pi(\frac{2}{3}r^3)-\pi(-\frac{2}{3}r^3)$

$V=\frac{4}{3}\pi r^3$

This is how the volume of a sphere can be derived, and similary, this is how we can derive the area of our segment:

$V=\pi(25^2x-\frac{x^3}{3})|_{-23}^{-3}$

$V=\pi(25^2((-3)-(-23))-\frac{(-3)^3-(-23)^3}{3})=\frac{25360}{3}\pi$

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Example Question #3973 : Calculus

Sphere segment

A sphere with a radius of 5 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 3 units to the left of the origin and the second cut is 4 units to the right of the origin, what is the volume of the segment?

Possible Answers:

$\frac{434}{3}\pi$

$\frac{76}{3}\pi$

$\frac{38}{3}\pi$

$\frac{152}{3}\pi$

$\frac{217}{3}\pi$

Correct answer:

$\frac{434}{3}\pi$

Explanation:

To approach this problem, begin by picturing how the sphere might be projected onto the two-dimensional Cartesian plane:

Sphere segment outlined

In its two-dimensional projection, we notice the outline of a circle. The formula for a circle of radius in the Cartesian coordinate system is given as:

x^2+y^2=r^2

Which can be rewritten in terms of as $y=\sqrt{r^2-x^2}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

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

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{a}^{b}\pi(r^2-x^2)dx$

$V=\pi(r^2x-\frac{x^3}{3})|_{a}^{b}$

Consider the points where circle contacts the x-axis, the greatest and smallest possible values of on the circle:

$V=\pi(r^2x-\frac{x^3}{3})|_{-r}^{r}$

$V=\pi(r^2r-\frac{r^3}{3})-\pi(r^2(-r)-\frac{(-r)^3}{3})$

$V=\pi(\frac{2}{3}r^3)-\pi(-\frac{2}{3}r^3)$

$V=\frac{4}{3}\pi r^3$

This is how the volume of a sphere can be derived, and similary, this is how we can derive the area of our segment:

$V=\pi(5^2x-\frac{x^3}{3})|_{-3}^{4}$

$V=\pi(5^2((4)-(-3))-\frac{(4)^3-(-3)^3}{3})=\frac{434}{3}\pi$

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Example Question #3974 : Calculus

Sphere segment

A sphere with a radius of 7.2 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 1.8 units to the left of the origin and the second cut is 3.6 units to the right of the origin, what is the volume of the segment?

Possible Answers:

836.7

824.5

240.4

881.2

255.9

Correct answer:

824.5

Explanation:

To approach this problem, begin by picturing how the sphere might be projected onto the two-dimensional Cartesian plane:

Sphere segment outlined

In its two-dimensional projection, we notice the outline of a circle. The formula for a circle of radius in the Cartesian coordinate system is given as:

x^2+y^2=r^2

Which can be rewritten in terms of as $y=\sqrt{r^2-x^2}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

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

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{a}^{b}\pi(r^2-x^2)dx$

$V=\pi(r^2x-\frac{x^3}{3})|_{a}^{b}$

Consider the points where circle contacts the x-axis, the greatest and smallest possible values of on the circle:

$V=\pi(r^2x-\frac{x^3}{3})|_{-r}^{r}$

$V=\pi(r^2r-\frac{r^3}{3})-\pi(r^2(-r)-\frac{(-r)^3}{3})$

$V=\pi(\frac{2}{3}r^3)-\pi(-\frac{2}{3}r^3)$

$V=\frac{4}{3}\pi r^3$

This is how the volume of a sphere can be derived, and similary, this is how we can derive the area of our segment:

$V=\pi(7.2^2x-\frac{x^3}{3})|_{-1.8}^{3.6}$

$V=\pi(7.2^2((3.6)-(-1.8))-\frac{(3.6)^3-(-1.8)^3}{3})=824.5$

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Example Question #3975 : Calculus

Using the Shell Method, find the volume of the solid obtained by rotating the region bounded by $y=(x-1)(x-3)^{2}$ and the x-axis, about the y-axis.

Possible Answers:

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

$V=\frac{24\pi}{5}$

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

$V=\frac{18\pi}{5}$

$V=6\pi$

Correct answer:

$V=\frac{24\pi}{5}$

Explanation:

The surface area of a cylinder formed by this region is $A(x)=2\pi(radius)(height)$ , where radius=x and $height = (x-1)(x-3)^{2}$ . This gives us $A(x)= 2\pi(x)((x-1)(x-3)^{2})=2\pi(x^{4}-7x^{3}+15x^{2}-9x)$ .

To obtain the volume, we integrate over this area, finding the bounds where $(x-1)(x-3)^{2}=0$ , at x=1 and x=3. Integrating, we get:

$V=\int_1^32\pi(x^{4}-7x^{3}+15x^{2}-9x)dx=2\pi\left [ \frac{x^{5}}{5}-\frac{7x^{4}}{4}+5x^{3}-\frac{9x^{2}}{2}\right ]_1^3=\frac{24\pi}{5}$

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Example Question #61 : How To Find Volume Of A Region

Find the volume of the solid obtained by rotating the region bounded by $y=\sqrt[3]{x}$ , x=8 , and the x-axis, about the y-axis.

Possible Answers:

$V=18\pi$

$V=\frac{48\pi}{5}$

$V=\frac{128\pi}{5}$

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

$V=\frac{96\pi}{5}$

Correct answer:

$V=\frac{96\pi}{5}$

Explanation:

We can relabel the curve $y=\sqrt[3]{x}$ as $x=y^{3}$ .

The radius of a single shell can be expressed as and the width can be expressed as $8-y^{3}$ , giving a cross sectional area of $A(y)=2\pi(8y-y^{4})$ .

To obtain the volume of solid, we integrate over each section. The first cylinder will cut into the solid at y=0 and the last cylinder will intersect at y=2 .

Integrating, we get:

$V=2\pi\int_0^2(8y-y^{4})dy= 2\pi\left [ (4y^2-\frac{y^{5}}{5}) \right ]_0^2=\frac{96\pi}{5}$

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Example Question #3977 : Calculus

Find the volume of the solid generated when the region in the first quadrant bounded by the graph of $y=9-x^{2}$ , the x-axis, and the y-axis is revolved around the line x=-1 .

Possible Answers:

$63\pi$

$V=\frac{96\pi}{5}$

$V=\frac{153\pi}{2}$

$V=\frac{76\pi}{3}$

$V=\frac{120\pi}{7}$

Correct answer:

$V=\frac{153\pi}{2}$

Explanation:

$V=2\pi\int_0^3((x+1)-(9-x^{2}))dx=2\pi\int_0^3(-x^{3}-x^{2}+9x+9)dx=2\pi \left [ (-\frac{x^{4}}{4}-\frac{x^{3}}{3}+\frac{9x^{2}}{2}+9x) \right ]_0^3=\frac{153\pi}{2}$

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Example Question #61 : Volume

Find the volume of the region bounded by $y=\sqrt{x}$ , y=2-x , and the x-axis, and rotated about the x-axis.

Possible Answers:

$V=\pi$

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

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

$V=\frac{3\pi}{4}$

$V=\frac{5\pi}{6}$

Correct answer:

$V=\frac{5\pi}{6}$

Explanation:

Using $x=y^{2}$ and x=2-y , we find that the graphs intersect at (1,1).

The sample cylindrical shell's height is $h=2-y-y^{2}$ and the radius is r=y .

Integrating, we get:

$V=2\pi\int_0^1(y(2-y-y^{2}))dy=2\pi\left [ y^{2}-\frac{y^{3}}{3}-\frac{y^{4}}{4}\right ]_0^1=\frac{5\pi}{6}$

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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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Calculus 1 : Volume

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2944 : Functions

Example Question #61 : How To Find Volume Of A Region

Example Question #3973 : Calculus

Example Question #3974 : Calculus

Example Question #3975 : Calculus

Example Question #61 : How To Find Volume Of A Region

Example Question #3977 : Calculus

Example Question #61 : Volume

Example Question #61 : Volume

Example Question #3981 : Calculus

All Calculus 1 Resources

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