Solid of revolution (should be simple)

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Homework Help Overview

The problem involves finding the volume of a solid formed by rotating the area between the curves y=e^x and y=sin x over the interval from 0 to π. The context is within the study of solids of revolution.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the volume and the correctness of the anti-derivative used in the integration process. There is an exploration of potential mistakes in the sign of the anti-derivative and the implications of the rotation axis.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's reasoning. Some guidance has been offered regarding the anti-derivative, and there is a recognition of the need to clarify the rotation axis.

Contextual Notes

Participants are working under the assumption that the area is to be revolved around the x-axis, and there is a reference to a specific answer that may not align with the calculations presented.

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 !
 

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