Just to make sure there's no typo. Is that supposed to be
?.
That z on the left. Is that correct?.
If , then
I'm asked to find the volume which is inside both the sphere and the paraboloid .
My attempt: The equation of the sphere : . So it's a sphere with radius and center in . And the paraboloid is easy to visualize.
.
My guess is goes from the equation of the paraboloid to the equation of the sphere. So from to . I just want to be sure.
the boundaries of it is wrong you have
you know that
change from the parabola to the sphere as you said
we need to present with respect to
let z'=z-1/2 I just move the solid 1/2 unit down the new equations
since you need the positive upper coordinate ...........(1)
take
the integral
if you use cylindrical it will be easier I think
Today a friend of mine told me the right answer is . I had shown him your (NCA) post, he copied it in his draft. He went to university and checked the result with a teacher who said it is wrong. My friend didn't understand all what the teacher said, he just came up with .
As I trust you and that your post makes sense to me, I wonder where's the error of the teacher. I know he used cylindrical coordinates, but I don't really know what he did. It seems that for him goes from (so the paraboloid) to , which I don't know what is it (it has no other choice than being the sphere, but as you pointed out, the paraboloid is inside it. I'm quite confused!)
Does someone has anything to say?
your teacher is right! sorry i forgot to add the volume of the upper half of the sphere, which is to so the correct answer is
as i said before the paraboloid is completely inside the sphere for so the plane divides your solid into two parts: a lower part, which is the paraboloid, and an upper part,
which is the upper half of your sphere. i forgot to add the volume of the upper part of your solid! you should now understand the limits of the integral in your teacher's solution as well.