Calculating Volume of Solid Using Triple Integral

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1. Use a triple integral to find the volume of the given solid.

The solid enclosed by the cylinder x^2 + z^2 = 4 and the planes y = -1 and y + z = 4


This looked like a cylindrical coordinate system to me, except for the fact that it is not cylindrical around the z-axis but the y-axis. I tried to fix this problem by "rotating" my coordinate axes so that my old z-axis would be my new x-axis, x would become y, and y would become z. I'm not sure if this is a valid approach or not.
 
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It makes no difference mathematically how we name the coordinate axes. The coordinate system ##(r,\theta,y)##, with
##x=rcos\theta##
##z=rsin\theta##,
is as valid a cylindrical coordinate system as the usual ##(r,\theta,z)##.

Now the volume element is ##dxdydz=rdrd\theta dy##. Note that because of the plane ##y+z=4## all of the integration limits in the volume calculation are not constants, one of them is a function of ##r## and ##\theta##.
 
Hilbert1 is absolutely correct. However, if you lack his sophistication, you can just rewrite the problem switching the y and z. They are just symbols, right?
 
Thank you both so much!
 
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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