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Calculus 1 : Volume

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Example Question #1 : Volume

Suppose I want to construct a cylindrical container. It costs 5 dollars per square foot to construct the two circular ends and 2 dollars per square foot for the rounded side. If I have a budget of 100 dollars, what's the maximum volume possible for this container?

Possible Answers:

$\sqrt{\frac{10\pi}{3}}$

$\frac{70}{6}\sqrt{33\pi}$

$\frac{70}{3}\sqrt{\frac{\pi}{3}}$

$V=\frac{50}{3}\sqrt{\frac{10}{3\pi}}$

$\frac{15}{2}\sqrt{\frac{10\pi}{3}}$

Correct answer:

$V=\frac{50}{3}\sqrt{\frac{10}{3\pi}}$

Explanation:

Write down equations representing the volume(V) and cost of the cylinder.

$V=\pi r^2h$

$100=5(2\pi r^2)+2(2\pi rh)=10\pi r^2 +4\pi rh$

We want to find the values of and that will maximize the volume. Before taking the derivative of the volume equation, let's eliminate by using the cost equation.

$h=\frac{100-10\pi r^2}{4\pi r}$

Plug this into the volume equation

$V=\pi r^2\left (\frac{100-10\pi r^2}{4\pi r} \right )=r\left (\frac{100-10\pi r^2}{4} \right )$

$V=\frac{100r-10\pi r^3}{4}$

So now is eliminated in the volume equation. Take the derivative with respect to , set it equal to zero, and solve for .

$\frac{dV}{dr}=\frac{100-30\pi r^2}{4}=0$

$r=\sqrt{\frac{10}{3\pi}}$

We can use this value to find

$h=\frac{100-10\pi \left (\sqrt{\frac{10}{3\pi}} \right )^2}{4\pi \sqrt{\frac{10}{3\pi}}}=\frac{100-\frac{100}{3}}{4\sqrt{\frac{10\pi}{3}}}=\sqrt{\frac{3}{10\pi}} \frac{200}{12}$

$h=\frac{50}{3}\sqrt{\frac{3}{10\pi}}$

Now that we found AND that maximizes the volume, we can find that maximum volume

$V=\pi r^2h=\pi\left (\sqrt{\frac{10}{3\pi}} \right )^2 \frac{50}{3}\sqrt{\frac{3}{10\pi}}$

$V=\frac{50}{3}\sqrt{\frac{10}{3\pi}}$

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Example Question #2 : Volume

We have the function $f(x)=\sqrt{x}$ and it is used to form a three dimensional figure by rotating it about the line $y=4$ . Find the volume of that figure from $x=0$ to $x=5$ .

Possible Answers:

100

Volume is infinite

432.129

103.269

43.874

Correct answer:

103.269

Explanation:

Imagine a test rectangle with a length of $4-\sqrt{x}$ and it is rotated around $y=4$ to form a cicular disk with area $\pi (4-\sqrt{x})^2$ . The disk has a thickness $\Delta x$ so that its volume is $\pi(4-\sqrt{x})^2 \Delta x$ . To find the total volume of the figure, turn this into an integral.

$Volume=\pi\int_{0}^{5}(4-\sqrt{x})^2 dx$

Perform the integration

$Volume=\pi\int_{0}^{5}(x-8\sqrt{x}+16) dx=\pi\left ( \frac{1}{2}(5)^2-\frac{16}{3}(5)^{\frac{3}{2}}+16(5) \right )$

$Volume=103.269$

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Example Question #3 : Volume

Consider a volume V of height H along some axis, which we will simply call h, such that 0 < h < H . We can express the figure as cross-sectional areas A(h) perpendicular to this h-axis. For example for a cone whose base has radius r, we can choose the axis to go through the point of the cone at h = 0 and then the cone is a stack of little circles of radius r h/H , so $A(h) = \pi r^2 h^2/H^2$ .

What is in terms of A(h) ?

Possible Answers:

$V = H \cdot \left( \lim_{h \to H} A(h) - \lim_{h \to 0} A(h) \right )$

$V = \int_0^H A(h) ~ dh$

$V = \frac{1}{2H} \int_{-H}^H ~A(h) ~ dh$

$V = \int \frac{dh}{dx} ~ A(h) ~ dx + C$

$V = H ~ \cdot ~\frac{A(H) + A(0)}{2}$

Correct answer:

$V = \int_0^H A(h) ~ dh$

Explanation:

The volume is approximated by the Riemann sum formed when we stack n layers atop each other with $\Delta h = H/n$ , h_k = H k / n

$V \approx \sum_{k=1}^n A(h_k) ~ \Delta h$

In the limit as n gets large, this expression becomes the integral:

$V = \int_0^H A(h) ~dh$

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Example Question #4 : Volume

What is the volume inside the bowl z(x, y) = x^2 + y^2 , 0 < z < 1 ? Hint: This is a solid of revolution about the z-axis with a radius of $\sqrt{x^2 + y^2}$ .

Possible Answers:

$\pi/2$

3/2

$\pi/3$

$\pi/4$

Correct answer:

$\pi/2$

Explanation:

Since z = r^2 we know that the volume element

$dV = A ~ dz = \pi ~ r^2 ~ dz = \pi ~z ~dz$ ,

and integrating gives:

$V = \int_0^1 \pi ~z~ dz = \pi \left[ \frac{z^2}{2} \right ]_0^1 = \pi \left( \frac{1}{2} - 0 \right ) = \frac{\pi}{2}$

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Example Question #5 : Regions

The following function:

f(x) = x + x^2

Is rotated around the -axis to create a three-dimensional shape. What is the volume of this object within the interval x = 0 to ?

Possible Answers:

$\frac{196}{9}\pi$

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

$\frac{256}{15}\pi$

$\frac{128}{7}\pi$

Correct answer:

$\frac{256}{15}\pi$

Explanation:

Note that for a given value of x in the function:

f(x) = x + x^2

The value of f(x) is the distance between the corresponding point on the curve and the x-axis. If the curve is rotated around the x-axis to create a three-dimensional object, f(x) can be seen as the radius of a circular cross-section of the object for any value of x.

To define the volume of this object, we can view it as a sum of infinitely thin disks stacked along the x-axis:

$V = \int \pi(f(x)^{2})dx$

Note how this follows the formula for the volume of a cylinder or disk

$( \pi \cdot height \cdot radius ^2 )$ ,

where f(x) is the radius and is the height.

Plugging in our function and range, we can rewrite this as:

$V = \int_{0}^{2} \pi(x+x^{2})^{2}dx$

$V = \int_{0}^{2}\pi(x^{4} + 2x^{3} + x^{2})$

Integrating this yields

$V = \pi \left(\frac{x^{5}}{5} + \frac{x^{4}}{2} + \frac{x^{3}}{3}\right)$

Using the upper bound of 2 and lower bound of 0 we find,

$V = \pi\left(\frac{2^{5}}{5}+\frac{2^{4}}{2} +\frac{2^{3}}{3}\right) - \pi\left(\frac{0^{5}}{5}+\frac{0^{4}}{2} +\frac{0^{3}}{3}\right) =\pi \frac{256}{15}$ .

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Example Question #4 : Volume

Determine the volume for the region bounded by y=x and y=x^2 if the region is revolved about the -axis.

Possible Answers:

$\frac{\pi}{30}$

$\frac{\pi}{5}$

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

$\frac{2\pi}{11}$

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

Correct answer:

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

Explanation:

In the region bounded by y=x and y=x^2 , write the formula to determine the volume of a solid revolved about an axis. This is the washer method.

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

Determine which function is the top function. This will be the big radius, and the small radius is represented by the bottom curve. The bounds of the integral are at the intersections of the two bounded functions. Evaluate the integral.

$V=\pi \int_{0}^{1} ((x)^2-(x^2)^2)\:dx$

$V=\pi \int_{0}^{1} x^2-x^4\:dx$

$V=\pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_{0}^{1}=\pi\left[\frac{1}{3}-\frac{1}{5}\right] = \frac{2\pi}{15}$

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Example Question #6 : How To Find Volume Of A Region

The unit circle is described by $y=\sqrt{1-x^2}$ and is graphed below. Find the volume of the shaded region rotated around the -axis. The shaded region starts at $x=-\sqrt{2}/2$ and ends at $x=\sqrt{2}/2$ .

Screen shot 2015 05 31 at 12.26.17 pm

Possible Answers:

$\frac{5\pi}{6}\sqrt{8}$

$\pi$

$\frac{5\pi}{6}\sqrt{2}$

$2\sqrt{2}\pi$

$\sqrt{2}\pi$

Correct answer:

$\frac{5\pi}{6}\sqrt{2}$

Explanation:

To find the volume of a region rotated around the -axis, we must find:

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

In our case, this is:

$V=\pi\int_{\sqrt{2}/2}^{\sqrt{2}/2}{(\sqrt{1-x^2})^2}dx$

$V=\pi\int_{\sqrt{2}/2}^{\sqrt{2}/2}{1-x^2} \; dx$

$V =\pi \left[x-\frac{1}{3}x^3\right]_{-\sqrt{2}/2}^{\sqrt{2}/2}$

$V =\pi \left(\sqrt{2}/2-\frac{1}{3}(\sqrt{2}/2)^3\right) - \pi\left(-\sqrt{2}/2-\frac{1}{3}(-\sqrt{2}/2)^3\right)$

$V = \pi\left(\frac{\sqrt{2}}{2}-\frac{1}{3}\cdot \frac{2\sqrt{2}}{8}\right) - \pi\left(-\frac{\sqrt{2}}{2}-\frac{1}{3}\left(-\frac{2\sqrt{2}}{8}\right) \right )\right)$

$V = \pi \left(\frac{\sqrt{2}}{2}-\frac{2\sqrt{2}}{24} \right )-\pi\left(-\frac{\sqrt{2}}{2} +\frac{2\sqrt{2}}{24}\right )$

$V = \pi \left(\frac{12\sqrt{2}}{24}-\frac{2\sqrt{2}}{24} \right )-\pi\left(-\frac{12\sqrt{2}}{24} +\frac{2\sqrt{2}}{24}\right )$

$V = \pi \left(\frac{10\sqrt{2}}{24} \right )-\pi\left(-\frac{10\sqrt{2}}{24} \right )$

$V = \frac{20\pi}{24}\sqrt{2}$

$V = \frac{5\pi}{6}\sqrt{2}$

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Example Question #7 : Regions

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

f(x)=x^2

revolved around the -axis on the interval [0,10] .

Possible Answers:

$20000\pi$ units cubed

$\frac{1000}{3}\pi$ units cubed

$10000\pi$ units cubed

$40000\pi$ units cubed

Correct answer:

$20000\pi$ 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,

$V=\pi\int_{0}^{10}(x^2)^2 dx=\pi\int_{0}^{10}x^4 dx$ .

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

$\pi\int_{0}^{10}x^4 dx=\pi\left(\frac{1}{5}x^5\right)\left.\begin{matrix} \, \end{matrix}\right|_{0}^{10}$ .

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

$\pi\left(\frac{1}{5}x^5\right)\left.\begin{matrix} \, \end{matrix}\right|_{0}^{10}=\pi \left(\left[\frac{1}{5}(10)^5\right]-\left[\frac{1}{5}(0)^5\right]\right)=20000\pi$ .

As such, the volume is

$20000\pi$ units cubed.

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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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Calculus 1 : Volume

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1 : Volume

Example Question #2 : Volume

Example Question #3 : Volume

Example Question #4 : Volume

Example Question #5 : Regions

Example Question #4 : Volume

Example Question #6 : How To Find Volume Of A Region

Example Question #7 : Regions

Example Question #8 : Regions

Example Question #5 : Volume

All Calculus 1 Resources

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