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bodensee9
Jan30-07, 11:11 AM
1. The problem statement, all variables and given/known data
Hello, I was wondering if someone could help me with the following. Supposed I am asked to find the volume bounded by the cylinders x^2+y^2=1 and the planes y = z, x = 0, z = 0 in the first octant.


2. Relevant equations
So this is what I tried to do. The boundaries should be: x is between 0 and 1 and y is between the squareroot of (1-x^2) and 0, or you can have y is between 0 and 1 and x is between the squareroot of (1-y^2) and 0. So wouldn't the double integral be the integral of

the squareroot of 1-x^2dydx, where you first evaluate it from 0 to the squareroot of (1-x^2), and then you evaluate it again from 0 to 1? Thanks!!!


3. The attempt at a solution

HallsofIvy
Jan30-07, 12:28 PM
Since you are in the first octant, yes, x runs between 0 and 1. For each x, then y runs from 0 up to the circle, y= \sqrt{1- x^2}. Finally, for each x and y, z runs from 0 up to the plane z= y. The volume is given by
[tex]\int_{x=0}^1\int_{y=0}^{\sqrt{1-x^2}}\int_{z=0}^y dzdydx= \int_{x=0}^1\int_{y=0}^{\sqrt{1-x^2}}y dydx[/itex]
No, that is NOT \sqrt{1- x^2}dydx! You don't get the square root until after integrating with respect to y- and then, since the integral of ydy will involve y2, you don't really have a square root to integrate with respect to x!

bodensee9
Jan30-07, 12:43 PM
Oh I see now! Thanks!!!