Calculus 1 : How to find volume of a region

Study concepts, example questions & explanations for Calculus 1

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Example Questions

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Example Question #1 : How To Find Volume Of A Region

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:

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

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

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

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

Example Question #1 : How To Find Volume Of A Region

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:

Volume is infinite

Correct answer:

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

Example Question #2 : Volume

Consider a volume V of height H along some axis, which we will simply call h, such that . 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  , so .

What is  in terms of ?

 

Possible Answers:

Correct answer:

Explanation:

The volume is approximated by the Riemann sum formed when we stack n layers atop each other with 

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

Example Question #3 : Volume

What is the volume inside the bowl ,  ? Hint: This is a solid of revolution about the z-axis with a radius of  .

Possible Answers:

Correct answer:

Explanation:

Since  we know that the volume element

,

and integrating gives:

Example Question #4 : Volume

The following function:

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

Possible Answers:

Correct answer:

Explanation:

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

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:

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

,

where  is the radius and  is the height.

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

or

Integrating this yields

 

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

.

 

Example Question #5 : Volume

Determine the volume for the region bounded by  and  if the region is revolved about the -axis.

Possible Answers:

Correct answer:

Explanation:

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

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.

 

Example Question #6 : Volume

The unit circle is described by  and is graphed below. Find the volume of the shaded region rotated around the -axis. The shaded region starts at  and ends at .

Screen shot 2015 05 31 at 12.26.17 pm

Possible Answers:

Correct answer:

Explanation:

The unit circle is described by  and is graphed below. Find the volume of the shaded region rotated around the -axis. The shaded region starts at  and ends at .

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

In our case, this is:

Example Question #7 : Volume

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

revolved around the -axis on the interval .

Possible Answers:

 units cubed

 units cubed

 units cubed

 units cubed

Correct answer:

 units cubed

Explanation:

The formula for the volume is given as

 

where .

As such,

.

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

.

Applying this rule we get 

 .

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

.

As such, the volume is

 units cubed.

Example Question #2 : How To Find Volume Of A Region

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

revolved around the -axis on the interval .

Possible Answers:

 units cubed

 units cubed

 units cubed

 units cubed

Correct answer:

 units cubed

Explanation:

The formula for the volume is given as

 

where .

As such,

 .

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

.

Applying this rule we get 

.

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

.

As such, the volume is

 units cubed.

Example Question #9 : Volume

Find the volume of the solid shape made by rotating  about the -axis on the interval .

Possible Answers:

Correct answer:

Explanation:

Find the volume of the solid shape made by rotating f(x) about the x-axis on the interval 

Recall the following formula for volume of a solid shape:

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

So we need to put it all together:

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

         

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

So our answer is:

 

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