L.+Hollifield

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__**Finding the volume of a region revolving around a given line.**__ Integration from 0 to π/2, π/2 to 3π/2, and 3π/2 to π, of the region bounded by the lines y=cos(x)+1, y=1, and x=2π. About the line y=1.

====// Great job Levi! I liked the way that you set the problems up in parts and then did 1 integration. There is a mistake in the 2nd part of the integration by parts. Do you see what you did? (Mrs. Self) //====

What was being found if he would not have went back and put the squares on the problem?
he would have found 2-D circles instead of wrapping the area around the axis -lovebk

// He would have found the area between the curve and the line y=1, instead of an volume. //(Mrs.Self)
== When integrating by parts (the first section of integrals, I get a volume of 69.546. Therefore if doing that by parts you do not get the answer that he got at the end. Is there something wrong with the problem? == on the second portion of the equation, Levi wrote 1-cosx+1 during the integration instead of cosx+1-1 because you are rotating around the line y=1 so you take 1 away from each equation in the integration. If you change that one little part then you get the correct answer of 9.869. ~Eve

Why does he set that equal to the integration of cos x from 0 to 2π? Why does this work?
During the integration, when accounting for the rotation around y=1 rather than the x-axis, you have to subtract 1 from y=cos(x)+1. When you do this in every integral, the 1 cancels out, and you end up with just y=cos(x).-Levi

Is there another way to do this problem?
if you were to rotate the area from the first intersection to the second around the line y=1 and double it, you would recieve the same area as you would if you solved the problem the way Levi did. -lovebk. // Great Job! (Mrs. Self) //