Question

HELP!!!!!!PLEASE!!!!!!!
Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Then use your calculator to evaluate the integral correct to five decimal places.
y=0, cos^2(x), -pi/2less than or equal to x less than or equal to pi/2.
a. about the x-axis
b. about y=1

Answer 1

Ok...

With volumes of revolution like this we can say that solid formed is made up of a bunch of either circles or washers, depending.

And we know the area of a circle (pi r^2) or of a washer (pi * (R^2-r^2)), so we can just find all of those and then multiply it by the height.

So in order to do this we can set up the integral:\[\int\limits_{a}^{b}\pi \times (R^{2}-r^{2})dx\](When it's a circle, r=0.)

So in the first one it's around y=0.

So how would you set up the integral for finding the area under the curve of (cos(x))^2 from -pi/2 to pi/2?

Answer 2

\[2\int\limits_{0}^{\Pi/2}\cos^2(x)dx\]

Answer 3

That works. So in this case, the inner radius doesn't exist (you have a circle), so you just need to multiply it by pi in order to find the volume of revolution.

Answer 4

Could you help me set up the second one of about y=1?

Answer 5

Certainly!

This one is a lot trickier though. In this one, we will have both an outer radius (R) and an inner radius (r).

Now, the outer radius is a bit easier to consider first in this case. Since we're rotating around y=1, of the two lines we have, y=0 is farther away at every point. To get this radius, we need to find the distance between y=1 and our function. This is just 1, since we find it by
radius=line of rotation - function
So R = 1-0=1

Then for R^2 it is just 1.

The inner radius can be found the same way; r=1-(cos(x))^2

\[r^{2}=(1-\cos^{2}x)^{2}\]


So then it's just \[\int\limits_{a}^{b}\pi \times (R^{2}-r^{2})dx\]


(As a note I did make a mistake earlier; I should have told you that you needed to square (cos(x))^2 and then multiply by pi, since (cos(x))^2 is your radius.)

Answer 6

thanks for the help!

Answer 7

You're welcome! I hope that made sense. :)

Answer 8

Btw, those are my initials...you're not me from the past are you?! -_-