Solid of revolution (should be simple)

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Paulo Serrano
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Hey. Thanks in advanced for the help. This site has helped me a lot through the years.

Homework Statement



Find the volume of the solid formed by rotating the area within y=e^x and y=sin x when 0<x<pi

Homework Equations





The Attempt at a Solution



I've tried it like 10 times on the whiteboard, and did it quickly on a sheet of paper so someone call tell me where I'm going wrong...the answer is supposedly pi/8(e^(2pi)-1)

Here's my work. http://min.us/mvfTL55
 
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The anti-derivative if cos(x) is +sin(x), not -sin(x) .
 
Are you sure that's where I made my mistake? The reason it became positive (I think) is not that I got the anti-derivative mixed up. It's because the negative sign outside the parenthesis canceled out the one before the pi*integral of cos2x/2
 
Of course! You're right.

What is the correct answer?

What axis is this area to be revolved around?
 
Around the x-axis. answer is pi/8(e^(2pi)-1)

It looks kinda similar...but not quite.
 
Your work looks good to me !
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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