Given function f(x,y,z)=y
Region bounded by x+y+z=2, x^2+z^2=1, and y=0
Integrate given function over indicated region.
Lets re-arrange, so in effect what we have is a plane and a cylinder of radius 1
$\displaystyle z=2-x-y$
$\displaystyle z= \sqrt{1-x^2} $
To find our domain in the XY plane let these equal eachother such that
$\displaystyle 2-x-y= \sqrt {1-x^2}$
$\displaystyle y= 2-x- \sqrt {1-x^2}$
Now, what we are looking for is
$\displaystyle \iiint ydV$
We know our Z bounds (between the 2 equations we formatted). But what about X and Y. Well we might want to evaluate Y last because we have a function Y as our density.
So let us find when x=0
$\displaystyle y= 2- \sqrt {1}$
So now we know Y runs from 0 to the above. Find your X bounds, and which Z is bigger (either the cylinder or the plane) and put in your bounds accordingly. We are done!
From that, I got that the limits of integration are:
z= [-sqrt(1-x^2) , sqrt(1-x^2)]
y= [0 , 2-x-sqrt(1-x^2)]
x= [-1,1]
The limits on z are from the cylinder equation, bounded on the y axis by the intersection of the cylinder and plane. Does this sound right? Then the limits on x are just from "end to end" of the cylinder.
Typically, when we have an integral such as
$\displaystyle \iint ydV$
It is easiar to evaluate y in terms of constant bounds (i.e integrate y last) but you don't have to. In this case it might be easiar to find the bounds by integrating x last. So let us do that.
$\displaystyle
y= 2-x- \sqrt {1-x^2}
$
Set y=0 and find your bounds for X. Then y of course will be from
$\displaystyle
0 \le y \le 2-x- \sqrt {1-x^2}
$
In the z plane you are bounded by both of the equations we derived. So in fact, our bounds would look something like
$\displaystyle \int_{ \sqrt{1-x^2}}^{2-x-y} dz$
Of course since I havent actually figured out the bounds for x, i am only guessing that
$\displaystyle \sqrt{1-x^2} \le 2-x-y$ for region R
Edit- it is 3am and this is the most I can muster for tonight, I will actually compute this tomorrow is somebody else hasent done so already.