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Calculus 1 : Calculus

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Example Questions

Example Question #8 : Regions

Using the method of cylindrical disks, find the volume of the region of the graph of

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

revolved around the -axis on the interval [0,3] .

Possible Answers:

$\frac{9}{4}\pi$ units cubed

$\frac{9}{2}\pi$ units cubed

${}4\pi$ units cubed

$9\pi$ units cubed

Correct answer:

$\frac{9}{2}\pi$ units cubed

Explanation:

The formula for the volume is given as

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

where r=f(x) .

As such,

$V=\pi\int_{0}^{3}(\sqrt{x})^2 dx$ .

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

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

And by the corollary of the first Fundamental Theorem of Calculus,

$\pi\left(\frac{1}{2}x^2\right)\left.\begin{matrix} \, \end{matrix}\right|_{0}^{3}= \pi\left(\left[\frac{1}{2}(3)^2\right]-\left[\frac{1}{2}(0)^2\right]\right)$ .

As such, the volume is

$\frac{9}{2}\pi$ units cubed.

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Example Question #5 : Volume

Find the volume of the solid shape made by rotating f(x) about the -axis on the interval [0,3] .

f(x)=x^4+6

Possible Answers:

$\frac{14391 \pi}{9}$

$\frac{14391 \pi}{5}$

$\frac{1439 \pi}{5}$

$\frac{14391}{5}$

Correct answer:

$\frac{14391 \pi}{5}$

Explanation:

Find the volume of the solid shape made by rotating f(x) about the x-axis on the interval [0,3]

f(x)=x^4+6

Recall the following formula for volume of a solid shape:

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

Where A(x) is equal to the the area of a disk made by rotating our function:

$A(x)=\pi r^2$

r=f(x)

So we need to put it all together:

$A(x)=\pi(x^4+6)^2=\pi(x^8+12x^4+36)$

Then, we know that our limits of integration must be 0 and 3, because they are the interval that we are working with.

$V=\pi \int_{0}^{3}x^8+12x^4+36dx$

$V=\pi \left(\frac{x^9}{9}+\frac{12x^5}{5}+36x\right) _0 ^3$

(Notice that plugging in 0 will yield 0, so we only need to really worry about the 3)

$V=\pi \left(\frac{3^9}{9}+\frac{12(3)^5}{5}+36(3)\right)=\frac{14391 \pi}{5}$

So our answer is:

$\frac{14391 \pi}{5}$

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Example Question #11 : How To Find Volume Of A Region

Find the volume of the solid generated by rotating the shape bounded between the functions y=9 and y=x^2 .

Possible Answers:

$\frac{243}{5}\pi$

$\frac{972}{5}\pi$

$\frac{1944}{5}\pi$

$\frac{3888}{5}\pi$

$\frac{486}{5}\pi$

Correct answer:

$\frac{1944}{5}\pi$

Explanation:

The first step is to find the lower and upper bounds, the points where the two functions intersect:

x^2=9

x_l=-3;x_u=3

Knowing these, the solid generated will have the following volume:

$V=\int_{-3}^{3}(\pi(9)^2-\pi(x^2)^2)dx=\int_{-3}^{3}(\pi(81)-\pi(x^4))dx=\pi\left(81x-\frac{x^5}{5}\right)|_{-3}^3$

$V=\pi\left(81(3)-\frac{243}{5})-(81(-3)+\frac{243}{3}\right)=\frac{1944}{5}\pi$

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Example Question #2892 : Functions

What is the volume of the solid formed when the function

$y=\frac{3x^2}{2}$

is revolved around the line y = 0 over the interval [0, 2] ?

Possible Answers:

$12\pi$

$4\pi$

$8\pi$

$2\pi$

Correct answer:

$4\pi$

Explanation:

The disc method can be used to find the volume of this solid of revolution. To use the disc method, we must set up and evaluate an integral of the form

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

where a and b are the endpoints of the interval, and R(x) is the distance between the function and the rotation axis.

First find R(x):

$R(x)=\frac{3x^2}{2}-0=\frac{3x^2}{2}$

Next set up and evaluate the integral:

$\pi \int_{0}^{2}\frac{3x^2}{2}dx=\pi \left(\frac{3(2)^3}{6}-\frac{3(0)^3}{6} \right)$

Simplifying, we find that the volume of the solid is

$\pi \left(\frac{3(8)}{6}-0 \right)=4\pi$

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Example Question #11 : Volume

Find the volume of the equation $y=x^2 \; \; \; 0\leq x \leq 2$ revolved about the -axis.

Possible Answers:

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

$4\pi$

$\frac{8\pi }{3}$

$8\pi$

Correct answer:

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

Explanation:

This problem can be solved using the Disk Method and the equation

$\int_{a}^{b}\pi [y(x)]^2dx \; \; \; a\leq x\leq b$ .

Using our equation to formulate this equation we get the following.

$\\ \int_{0}^{2}\ \pi x^4dx \\ \\$

Applying the power rule of integrals which states

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

we get,

$\\ =\pi \frac{x^{4+1}}{4+1}\\ \\ =\frac{\pi x^5}{5}|_0^2\\ \\=\frac{32\pi }{5}$ .

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Example Question #13 : How To Find Volume Of A Region

Find the volume of the area between $f(x)=\sqrt{x}$ and g(x)=x^3 .

Possible Answers:

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

$\frac{-5\pi }{14}$

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

$\frac{5\pi }{14}$

Correct answer:

$\frac{5\pi }{14}$

Explanation:

Graph1

Looking at the graph, we see that the curves intersect at the point (1,1) and $f(x)=\sqrt{x}$ is on top.

Using the general equation

$\int_{a}^{b}\pi [(f(x))^2-(g(x))^2]dx \; \; \; a\leq x\leq b$ where f(x) is on top.

This gives us that the volume is

$\int_{0}^{1}\pi [(\sqrt{x})^2-(x^3)^2]dx=\int_{0}^{1}\pi (x-x^6)dx =\frac{5\pi }{14}$

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Example Question #14 : How To Find Volume Of A Region

Find the volume of the region enclosed by y=x^3 and $y=\sqrt{x}$ rotated about the line x=1 .

Possible Answers:

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

$\frac{13\pi}{30}$

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

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

Correct answer:

$\frac{13\pi}{30}$

Explanation:

The volume can be solved by

$dV=\pi(R^2-r^2)$ where R=1-y^2 and $r=1-\sqrt[3]{y}$ from 0 to 1 where the curves intersect.

Then, we formulate

$V=\int_{0}^{1}\pi[(1-y^2)^2-(1-\sqrt[3]{y})^2]dy=\frac{13\pi }{30}$

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Example Question #15 : How To Find Volume Of A Region

What is the volume of the solid formed by revolving the curve $y=\sqrt{x}$ about the x-axis on $x\in(0,4)$ ?

Possible Answers:

$8\pi$

$2\pi$

$6\pi$

$\pi$

$4\pi$

Correct answer:

$8\pi$

Explanation:

We take the region and think about revolving rectangles of height $\sqrt{x}$ and width $\Delta x$ about the line y=0 . This forms discs of volume $\pi(\sqrt{x})^2\Delta x$ . We then sum all of these discs on the interval and take the limit as the number of discs on the interval approaches infinity to arrive at the definite integral $\pi\int_{0}^{4}(\sqrt{x})^2dx=\pi\int_{0}^{4}xdx=\pi(\frac{x^{2}}{2})|_{0}^{4}=\pi(4^{2}/2)=8\pi$

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Example Question #16 : How To Find Volume Of A Region

For any given geometric equation with one variable, volume over a given region is defined as the definite integral of the surface area over that specific region.

Given the equation for surface area of any sphere, $A(r)=4\pi*r^2$ , determine the volume of the piece of the sphere from r=1 to r=5 .

Possible Answers:

$\frac{256\pi}{3}$

$\frac{496\pi}{3}$

$\frac{500\pi}{3}$

$200\pi$

Correct answer:

$\frac{496\pi}{3}$

Explanation:

Recall that volume is the definite integral of surface area over a given region.

$V(r)=\int_{a}^{b} A(r)dr$

Given our surface area formula, we can determine the volume in one step.

$V(r)=\int_{1}^{5}4\pi r^2 dr= \frac{4}{3}\pi* 5^3- \frac{4}{3}\pi* 1^3= \frac{496\pi}{3}$

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Example Question #17 : How To Find Volume Of A Region

Find the volume of the solid generated by rotating the curve f(x)=3x^2+2x , for the range x=[0,2] around the x-axis.

Possible Answers:

$\frac{436\pi}{15}$

$\frac{1744\pi}{15}$

$12\pi$

$\frac{1744}{15}$

Correct answer:

$\frac{1744\pi}{15}$

Explanation:

To solve this problem, treat the function

f(x)=3x^2+2x

As the radius of a disc with infinitesimal thickness . This disc would then have a volume defined the surface area:

$\pi r^2:\pi(3x^2+2x)^2$

$\pi(9x^4+12x^3+4x^2)$

And a thickness or height:

The volume of the solid generated would be the sum of these discs, i.e. the integral:

$V=\pi \int_{0}^{2}(9x^4+12x^3+4x^2)dx=\pi(\frac{9x^5}{5}+3x^4+\frac{4x^3}{3})|_0^2$

$V=\pi(\frac{288}{5}+48+\frac{32}{3})-(0+0+0)$

$V=\frac{1744\pi}{15}$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #8 : Regions

Example Question #5 : Volume

Example Question #11 : How To Find Volume Of A Region

Example Question #2892 : Functions

Example Question #11 : Volume

Example Question #13 : How To Find Volume Of A Region

Example Question #14 : How To Find Volume Of A Region

Example Question #15 : How To Find Volume Of A Region

Example Question #16 : How To Find Volume Of A Region

Example Question #17 : How To Find Volume Of A Region

All Calculus 1 Resources

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