Determining regions to double/triple integrate over

[dr. dre]

I have two questions:

FIRST ONE:
Let D= {(x,y)|2<= x <=pi, 2<= x*y <=3}.

How would I find the region for y? Would i simply just divide the outside by x?


SECOND ONE:
I am asked to compute the volume over a solid via triple integral.
I am given W={(x,y,z)|1 <= x^2 + y^2 + z^2 <= e^2, z<=0}. I am told to compute the volume by making change of variables. Can anyone help me on starting out, especially finding the inequalities for x, y, and z?

[HallsofIvy]

"Let D= {(x,y)|2<= x <=pi, 2<= x*y <=3}.

How would I find the region for y? Would i simply just divide the outside by x?"

Yes, as long as x is positive (and it is between 2 and pi!), 2<= xy<=3 is the same as 2/x<= y<= 3/x. Take your outer integral from x= 2 to x= pi and your inner integral from y= 2/x to y= 3/x.

"I am asked to compute the volume over a solid via triple integral.
I am given W={(x,y,z)|1 <= x^2 + y^2 + z^2 <= e^2, z<=0}. I am told to compute the volume by making change of variables. Can anyone help me on starting out, especially finding the inequalities for x, y, and z?"

1= x^2+ y^2+ z^2 is a sphere with center (0,0,0) and radius 1. x^2+ y^2+ z^2= e^2 is a sphere with center (0,0,0) and radius e. This region is the upper half of the region between the two spheres. I suspect that the "change of variables" referred to is to spherical coordinates. rho, the radius variable, will go from 1 to e, theta, the "longitude", will complete the full circle, from 0 to 2pi, and phi, the "co-latitude", will go from 0 to pi/2.

Actually, if you know the formula for volume of a sphere, you don't need to do an integration at all! The volume is (2/3)pi(e^3-1). Since you are told to do a triple integral, use that as a check.