Question

CALC 3!!! fan, medal and best answer
Find the volume below the surface z=(sqrt(16-x^2-y^2)) which lies above the cone Z^2=3x^2+3y^2

Answer 1

first find the volume of the sphere then subtract the volume of the cone, each of these should have there own double integral.

Answer 2

what would you do to turn this into spherical coordinates

Answer 3

rho would be 4 theta and phi would be free, so you'd integrate dphi dtheta. this is because the first equation can be expressed as z^2 +y^2 +x^2=4^2

Answer 4

wait, so what would the integrals be?

Answer 5

in which coordinate style?

Answer 6

dxdy

Answer 7

oops i mean p2sinphidphidtheta

Answer 8

\[\int\limits_{-4}^{4}\int\limits_{-4}^{4}\sqrt{16-x ^{2}-y ^{2}}dxdy-\int\limits_{-\sqrt(8)}^{\sqrt(8)}\int\limits_{-\sqrt(8)}^{\sqrt(8)} \sqrt{3x^2+3y^2}dxdy\] the sqrt(8) and -Sqrt(8) can be found by setting sqrt(3x^2 + 3y^2)=sqrt(16-x^2-y^2)

Answer 9

but wouldnt it be easier with spherical? How would you do this problem with spherical coordinates?

Answer 10

The first integral would be \[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}4\rho^2\sin \phi d \phi d \Theta\] but the second integral is actually easier to leave in rectangular coordinates.

Answer 11

I just realized my boundaries are off on the first integrals i typed, the first integral should both be from -sqrt8 to sqrt8, the one up there actually takes all of the volume outside the cone not just the volume above it.

Answer 12

the spherical one is off to the upper coordinate of the phi integral should be .7559

Answer 13

Oh ok. but when I solved it i got
∫(0,pi/3)∫(0,4)∫(0,2pi) p^2 sinphi dtheta dp dphi

Answer 14

is that not even close?

Answer 15

I curious about what you did, I see you've got a triple integral which isn't really useful for three space. Phi never exceeds pi as it is the angle from the z axis, the integral from 0-4 drho would yield the volume of the entire sphere, not just that above the cone and o-pi/3 dtheta is a small wedge of that.

Answer 16

For this problem, my teacher said to use three space. By the way, the phi doesn't exceed theta since it goes from 0 to pi/3 while theta goes from 0 to 2pi.

Answer 17

You're right about the phi theta bounds, my mistake. A double integral is an analysis of volume in three space much like a single integral is an analysis of two space. The double integral is the way to go here.

Answer 18

oh hmm. But when you turn something into spherical doesnt it automatically need three integrals?

Answer 19

cuz the there would be phi, theta and p?

Answer 20

I took calc 3 last fall, I'm sorry; you are right your initial integral may also have been correct, again it is my fault.

Answer 21

No no its ok. I usually don't remember stuff after i take a course. Haha.

Answer 22

I think all you need from your above integral is a 16(rho)cos(phi), which is 16z as z=rho cos(phi).

Answer 23

is that what i would be integrating?

Answer 24

\[\int\limits_{0}^{\pi/3}\int\limits_{0}^{2\pi}\int\limits_{0}^{4}\rho^2\sin \Phi 16\rho \cos(\phi) d \rho d\]

Answer 25

the ending there should read \[d \rho d \Theta d \Phi \]

Answer 26

Ok, i understand. Finally, solving which i think i can do.