Question

Surface Integral z=5-sqrt(x^2+y^2) above the z=0 plane, integrate (x^2+y^2+5z^2)

Answer 1

@Kainui

Answer 2

\[\int\limits_{0}^{5}\int\limits_{0}^{\sqrt{25-y^2}} (x^2+y^2+5(5-\sqrt{x^2+y^2})^2)*\sqrt((-x/(x^2+y^2)^2 + (-y/(x^2+y^2))^2 +1))\]

Answer 3

wow that is messy to look at. but that's what i got

Answer 4

Lol, I'll give a hint. Do the parametricizing of the surface. Do all the cross product stuff. Find the region of the cone, which is easy.

Try using spherical coordinates for your integration.

Answer 5

Ugh, I mean polar coordinates. This is a double integral.

Answer 6

instead of doing the cross product, I just use the formula sqrt(dz/dx^2 + dz/dy^2 +1)

Answer 7

So I turned z=x^2+y^2+5z^2 into z=x^2+y^2+5(5-sqrt(x^2+y^2))^2

Answer 8

from that, I was able to get the bounds of 0 to 5 for y and 0 to sqrt(25-y^2) for x

Answer 9

I am definitely on board with @iPwnBunnies, find out what the cone part looks like and it'll be a lot simpler to look at and write down.

Answer 10

I am awful with polar coordinates

Answer 11

Lol, I find it easy to convert. We already know the cone has a circular region with radius 5. That should be plenty...I am now out. Peace Kainui and Mgb.

Answer 12

Thanks @iPwnBunnies much appreciate the help

Answer 13

so I got, z=r^2+5(5-r)

Answer 14

i mean for the function after plugging z in

Answer 15

\[\int\limits_{0}^{2*\pi}\int\limits_{0}^{5} (r^2-5r+25) *r \] dr dtheta

Answer 16

but im missing the sqrt(dz/dx^2 + dz/dy^2 +1) part still

Answer 17

i got sqrt(2) for that part

Answer 18

Well instead of using that, I would use the cross product version. That I believe is specifically rectangular coordinates and won't work here.

Answer 19

is the other part correct?

Answer 20

I'm sort of not following your way of doing it to be honest, I'm not sure it's correct until you get an answer in the end for me to compare it to.

Answer 21

I'm using:\[T(r, \theta) = <x,y, 5-\sqrt{x^2+y^2}>\]\[T(r, \theta )=<r \cos \theta, r \sin \theta, 5-r>\]

Answer 22

ok so then, \[for r, we get <\cos \theta, \sin \theta,-1>\]

Answer 23

for theta, \[<-rsin \theta,rcos \theta,0>\]

Answer 24

?

Answer 25

then cross those?

Answer 26

You got it. =)

Answer 27

alright so we get \[(r \cos(\theta), rsin(\theta),rcos^2(\theta)+rsin^2(\theta))\]

Answer 28

or <rcos(t), rsin(t), r> right?

Answer 29

Yes, you got it. I'm afraid I have to leave, I've been distracted this whole time but I've got to go I'm sorry!

Answer 30

the magnetitude would be\[\sqrt((r \cos(x))^2+(r \sin(x))^2+(r)^2)\]

Answer 31

aww ok thanks for the help

Answer 32

@mathmale help me! haha