Volume by Cross Section Verification

• Mar 30th 2009, 09:00 PM
Moderatelyconfused
Volume by Cross Section Verification
I had to make a model of the function y=sin(x) and then find the volume of that model.
Below is a picture of the model I made, and vertical cardboard sections were placed 1/2 an inch apart from each other so (pi/2) = 2.5 inches.

http://i40.tinypic.com/zn9e75.jpg

I know that I need to integrate the area formula of the cross section with the limits to find area in in^2.

The volume of each square cross section is (sin(x)^2), and the model goes from -pi to pi.

I'm pretty sure the indefinite integral of this is -cos^2(x), but
I don't know where to take this problem from here. When I did the math in my calculator, I got pi in^3 as the answer, but by no means am I positive.

• Mar 30th 2009, 09:05 PM
mollymcf2009
Quote:

Originally Posted by Moderatelyconfused
I had to make a model of the function y=sin(x) and then find the volume of that model.
Below is a picture of the model I made, and vertical cardboard sections were placed 1/2 an inch apart from each other so (pi/2) = 2.5 inches.

http://i40.tinypic.com/zn9e75.jpg

I know that I need to integrate the area formula of the cross section with the limits to find area in in^2.

The volume of each square cross section is (sin(x)^2), and the model goes from -pi to pi.

I'm pretty sure the indefinite integral of this is -cos^2(x), but
I don't know where to take this problem from here. When I did the math in my calculator, I got pi in^3 as the answer, but by no means am I positive.

WOW!! That is REALLY cool!!

When you are finding the volume for a trig function that is identical on both sides, like sin(x) or cos(x) just find the volume of half of the interval i.e. $\displaystyle (0,\pi)$ and then multiply your answer by 2. That is the easier way to do that. Does that help you?
• Mar 30th 2009, 09:15 PM
Moderatelyconfused
Quote:

Originally Posted by mollymcf2009
WOW!! That is REALLY cool!!

When you are finding the volume for a trig function that is identical on both sides, like sin(x) or cos(x) just find the volume of half of the interval i.e. $\displaystyle (0,\pi)$ and then multiply your answer by 2. That is the easier way to do that. Does that help you?

Okay, I'm hoping that my integration was correct, but I'm not particurlarly confident in my calc skills.

I can see how separating the integral would make it easier, but I'm not really sure how to evaluage the integral to get an exact answer.

So once an exact answer is reached, does that mean that that answer is the actual volume of the cross section or is there an additional step?