Find the Volume (Solid of Revolution)

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Homework Help Overview

The problem involves finding the volume of a solid formed by revolving the area between the curves y = sin(x) and y = cos(x) around the x-axis, specifically between the limits x = 0 and x = π/4.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need for a complete problem statement, including the axis of revolution and the boundaries for integration. There are questions about how to determine the limits of integration and the reasoning behind the choice of the lower boundary at 0.

Discussion Status

Some participants have provided guidance on the need for a clearer description of the region to be revolved and suggested using graphical representations. There is an exploration of the method of washers for calculating the volume, but no consensus has been reached on the exact approach or interpretation of the problem.

Contextual Notes

Participants note that the original poster's understanding of the problem setup may be incomplete, and there is an emphasis on the importance of visualizing the region and the cross-section of the volume of revolution.

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Find the volume of the solid y = sinx, y=cosx, and x= pi/4, revolving around x-axis
I didn't really get this at all... do I plug pi/4 for x in y=sinx, y=cosx to get the integra boundaries?
 
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You need to give us the entire problem, including the axis this region has been revolved around.

After that, show us what you have tried. If you're stuck, your book most likely has some similar examples that show how to calculate volumes of revolution by cylindrical shells or by washers.
 
Mark44 said:
You need to give us the entire problem, including the axis this region has been revolved around.

After that, show us what you have tried. If you're stuck, your book most likely has some similar examples that show how to calculate volumes of revolution by cylindrical shells or by washers.

Sorry, forgot to mention it's revolving around the x-axis. Anyways, I got this from a friend pi * Int[0,pi/4] (cosx)^2 - (sinx)^2 dx), and I was wondering where the lower boundary, 0, came from.
 
The description of the region that will be revolved around the x-axis is not as complete as I would like. I believe it is the region bounded by the graphs of y = sin x and y = cos x between x = 0 (the y-axis) and the line x = pi/4.

If this is the right description, the region is sort of triangular, but with two curved sides. The "base" of this region runs along the y-axis between 0 and 1, and the two curves intersect at (pi/4, sqrt(2)/2).

You should have at least one graph: one showing the region to be revolved, and ideally, another that shows a cross-section of the volume of revolution. Your friend is using washers - disks with a hole in the middle.

The area of a washer is pi*(R^2 - r^2), where R is the larger radius and r is the smaller radius.

The volume of a washer is the area time the thickness, which is pi*(R^2 - r^2)*thickness, which can be either dx (vertical washers) or dy (horizontal washers).

For the limits of integration, figure out where the washers run. Vertical washers run left to right along the interval in question. Horizontal washers run bottom to top along the interval in question.

Can you get started with that?
 

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