Calculus 1 : Volume

Study concepts, example questions & explanations for Calculus 1

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Example Questions

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 \(\displaystyle y=9\) and \(\displaystyle y=x^2\).

Possible Answers:

\(\displaystyle \frac{3888}{5}\pi\)

\(\displaystyle \frac{243}{5}\pi\)

\(\displaystyle \frac{972}{5}\pi\)

\(\displaystyle \frac{486}{5}\pi\)

\(\displaystyle \frac{1944}{5}\pi\)

Correct answer:

\(\displaystyle \frac{1944}{5}\pi\)

Explanation:

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

\(\displaystyle x^2=9\)

\(\displaystyle x_l=-3;x_u=3\)

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

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

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

 

Example Question #2891 : Functions

What is the volume of the solid formed when the function 

\(\displaystyle y=\frac{3x^2}{2}\) 

is revolved around the line \(\displaystyle y = 0\) over the interval \(\displaystyle [0, 2]\)?

Possible Answers:

\(\displaystyle 12\pi\)

\(\displaystyle 8\pi\)

\(\displaystyle 4\pi\)

\(\displaystyle 2\pi\)

Correct answer:

\(\displaystyle 4\pi\)

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

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

where 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):

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

Next set up and evaluate the integral:

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

Simplifying, we find that the volume of the solid is

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

Example Question #11 : Volume

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

Possible Answers:

\(\displaystyle 4\pi\)

\(\displaystyle 8\pi\)

\(\displaystyle \frac{8\pi }{3}\)

\(\displaystyle \frac{32\pi }{5}\)

Correct answer:

\(\displaystyle \frac{32\pi }{5}\)

Explanation:

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

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

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

 \(\displaystyle \\ \int_{0}^{2}\ \pi x^4dx \\ \\\)

Applying the power rule of integrals which states 

\(\displaystyle \int x^n=\frac{x^{n+1}}{n+1}\)

we get,

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

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

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

Possible Answers:

\(\displaystyle \frac{-5\pi }{14}\)

\(\displaystyle \frac{5\pi }{12}\)

\(\displaystyle \frac{5\pi }{14}\)

\(\displaystyle \frac{-5\pi }{12}\)

Correct answer:

\(\displaystyle \frac{5\pi }{14}\)

Explanation:

Graph1

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

Using the general equation 

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

This gives us that the volume is  

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

Example Question #3922 : Calculus

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

Possible Answers:

\(\displaystyle \frac{12\pi}{5}\)

\(\displaystyle \frac{-2\pi}{5}\)

\(\displaystyle \frac{13\pi}{30}\)

\(\displaystyle \frac{2\pi}{5}\)

Correct answer:

\(\displaystyle \frac{13\pi}{30}\)

Explanation:

The volume can be solved by 

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

Then, we formulate 

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

Example Question #3921 : Calculus

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

Possible Answers:

\(\displaystyle 8\pi\)

\(\displaystyle 2\pi\)

\(\displaystyle \pi\)

\(\displaystyle 6\pi\)

\(\displaystyle 4\pi\)

Correct answer:

\(\displaystyle 8\pi\)

Explanation:

We take the region and think about revolving rectangles of height \(\displaystyle \sqrt{x}\) and width \(\displaystyle \Delta x\) about the line \(\displaystyle y=0\). This forms discs of volume \(\displaystyle \pi(\sqrt{x})^2\Delta x\). 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 \(\displaystyle \pi\int_{0}^{4}(\sqrt{x})^2dx=\pi\int_{0}^{4}xdx=\pi(\frac{x^{2}}{2})|_{0}^{4}=\pi(4^{2}/2)=8\pi\)

Example Question #12 : Volume

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, \(\displaystyle A(r)=4\pi*r^2\), determine the volume of the piece of the sphere from \(\displaystyle r=1\) to \(\displaystyle r=5\)

Possible Answers:

\(\displaystyle 200\pi\)

\(\displaystyle \frac{496\pi}{3}\)

\(\displaystyle \frac{500\pi}{3}\)

\(\displaystyle \frac{256\pi}{3}\)

Correct answer:

\(\displaystyle \frac{496\pi}{3}\)

Explanation:

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

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

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

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

 

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

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

Possible Answers:

\(\displaystyle \frac{436\pi}{15}\)

\(\displaystyle 12\)

\(\displaystyle \frac{1744\pi}{15}\)

\(\displaystyle 12\pi\)

\(\displaystyle \frac{1744}{15}\)

Correct answer:

\(\displaystyle \frac{1744\pi}{15}\)

Explanation:

To solve this problem, treat the function

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

As the radius of a disc with infinitesimal thickness \(\displaystyle dx\). This disc would then have a volume defined the surface area:

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

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

And a thickness or height:

\(\displaystyle dx\)

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

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

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

\(\displaystyle V=\frac{1744\pi}{15}\)

Example Question #3931 : Calculus

Find the volume generated by rotating the region between \(\displaystyle f(x)=x^2\) and \(\displaystyle g(x)=2x\) around the \(\displaystyle y\)-axis.

Possible Answers:

\(\displaystyle \frac{8}{3}\pi\)

\(\displaystyle \frac{8}{15}\pi\)

\(\displaystyle \frac{8}{15}\)

\(\displaystyle \frac{4}{3}\pi\)

\(\displaystyle \frac{8}{3}\)

Correct answer:

\(\displaystyle \frac{8}{3}\pi\)

Explanation:

Since the functions are being rotated around the y-axis, they should be rewritten. It may help to write them in terms of y:

\(\displaystyle y=x^2\)

\(\displaystyle h(y)=\sqrt{y}\)

And

\(\displaystyle y=2x\)

\(\displaystyle j(y)=\frac{y}{2}\)

Here, function designations \(\displaystyle [h(y),j(y)]\) have been used solely to eliminate the ambiguity of using x for both functions.

Now, there are two points of intersection for these functions:

\(\displaystyle \sqrt y= \frac{y}{2}:y=0,4\)

These will be the bounds for the integration to follow.

Finally, treat these two functions as the radii for two regions: a disk from the larger radius, and a hole from the smaller radius, allowing for the formation of a ring with a cross-sectional area:

\(\displaystyle A=\pi(\sqrt y^2 - \frac{y^2}{4})\)

These disks have infinitesimal thickness \(\displaystyle dy\), and so the volume of the solid can be found by adding all of these disks together; i.e. by taking the integral:

\(\displaystyle V=\int Ady=\int_{0}^{4}\pi(y-\frac{y^2}{4})dy\)

\(\displaystyle V=\pi(\frac{6y^2-y^3}{12})|_0^4\)

\(\displaystyle V=\frac{8}{3}\pi\)

 

 

Example Question #11 : Regions

Find the volume of the function

\(\displaystyle f(x)=5\) 

revolved around the \(\displaystyle x\)-axis on the interval \(\displaystyle [0, 10]\).

Possible Answers:

\(\displaystyle 210\pi\) cubic units

\(\displaystyle 250\pi\) cubic units

\(\displaystyle 125\pi\) cubic units

\(\displaystyle 500\pi\) cubic units

Correct answer:

\(\displaystyle 250\pi\) cubic units

Explanation:

The formula for the volume of the region revolved around the x-axis on the interval \(\displaystyle [a,b]\) is given as

 \(\displaystyle V=\pi \, \int_{a}^{b} r^2 \, dx\)

where \(\displaystyle r=f(x)\).

As such,

 \(\displaystyle V=\pi \, \int_{0}^{10} 5^2 \, dx\).

When taking the integral, we will use the inverse power rule which states,

 \(\displaystyle \int x^n = \frac{x^{n+1}}{n+1}\).

Applying this rule we get

\(\displaystyle \pi \, \int_{0}^{10} 5^2 \, dx=\pi \, \int_{0}^{10} 25 \, dx\)

 \(\displaystyle =\pi (25x)|_{0}^{10}\).

And by the corollary of the First Fundamental Theorem of Calculus

\(\displaystyle \pi (25x)|_{0}^{10}=\pi[25(10)-25(0)]=250\pi\).

As such, the volume is

\(\displaystyle 250\pi\) cubic units.

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