Question

Use spherical coordinates to evaluate the integral

Answer 1

\[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-y^2}} \int\limits_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}}y^2\sqrt{x^2+y^2+z^2}dzdxdy\]

Answer 2

@Callisto

Answer 3

ALright so we can notice a couple of things right off the bat

Answer 4

Like that for the dz we can see that the equation in regular coordinates is (z^2) + (y^2) + (x^2) = 4

Answer 5

But we know that the z squared, y squared, x squared is equal to rho

Answer 6

Oops yeah my bad I meant that

Answer 7

Thus we get that rho squared is equal to 4 and thus rho is equal to 2 (techincally plus or minus 2 but in spherical coordinates rho I believe describes the radius of the sphere assuming we can a sphere that we are finding the volume of)

Answer 8

Which leads me to believe that there cannot be a negative radius

Answer 9

Before I go on any further....what seems to be giving you the most trouble?

Answer 10

setting the bounds

Answer 11

Ok so we know that the boundaries of the integrals are for rho, theta, and phi

Answer 12

Rho I just explained....theta describing the rotation along the x y plane and phi describing the rotation along the z axis

Answer 13

Like if you imagine phi being a line that is starting from the positive z axis line and going down to the negative z axis line in a circular form

Answer 14

Well before I go on any further....based on what I have said...What do you think would be the boundaries for the dz intagral ?

Answer 15

0 to 2

Answer 16

Yes I believe that would be the answer for the first integral

Answer 17

Or rather the boundaries or limits for the first integral

Answer 18

Mind telling me what x, y and z are equal to in spherical coordinates?

Answer 19

Like the first I think is x = rho sin phi cos theta?

Answer 20

i have a question from the original problem: since limit of z from 4+x^2 +y^2 to 4 +x^2 +y^2 , the integral is not equal 0 at the beginning?

Answer 21

The limits would be equal to z

Answer 22

For the dz part I mean