Question

@ganeshie8

Answer 1

\[\huge \int\limits\limits_{-a}^{a} \int\limits\limits_{-\sqrt{a^2-y^2}}^{\sqrt{a^2-y^2}} \int\limits\limits_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\] converting this to spherical coordinates

Answer 2

looks like a sphere

Answer 3

is it `dzdxdy`

Answer 4

So I see that, \[z = \sqrt{a^2-x^2-y^2} \implies z^2 = a^2-x^2-y^2 \implies a^2=x^2+y^2+z^2\]

Answer 5

Yeah haha, i forgot to put rest of the integral, but w e just need help with bounds

Answer 6

p = a, 0<= p < = a

Answer 7

What about the other bounds >.<

Answer 8

you're done once you see that it is a sphere

Answer 9

p : 0->a
theta : 0->2pi
phi : 0->pi

Answer 10

Ah ok, I see now, thanks :P thought it was a bit more complicated then that

Answer 11

Is there a way to convert it as I did for p, without knowing it's a sphere

Answer 12

we can make it complicated by pretending that x^2+y^2+z^2 = a^2 is not a sphere

Answer 13

x = sqrt(a^2 - y^2)

x^2 + y^2 = a^2

Answer 14

thats a circle, it cannot be anything else...
similarly if we look at y bounds : -a -> a, it is ranging from left edge to right edge of "sphere "

Answer 15

it is good to use the knowledge that it is a sphere instead of relying fully on algebra..

Answer 16

Ok cool, I see it now, I knew it was a sphere, but I also wanted to know it algebraically, because I think that's what the question may be trying to teach, but knowing it's a sphere helps to haha, thanks!