Question

little help here?

Answer 1

i get that the equation to integrate becomes 1/rho, but not sure how to change the bounds exactly.

Answer 2

what is happening with those bounds....

Answer 3

i guess i have to convert them to spherical, but not sure how. i only have formulas for what x, y, and z are.

Answer 4

calc 3?

Answer 5

yes

Answer 6

doing calc 3 myself, i haven't seen this before but i want to point some things out

Answer 7

imagine putting paranthases before the third integrand and after the dy

Answer 8

gah... chain rule

Answer 9

OH OH!

Answer 10

i think it's simpler than that. just converting those bounds to a spherical bound. the equation is easy to convert.

Answer 11

no no no tell me if im wrong but if we make y equal to \[\sqrt{25-x^2-z^2}\]

Answer 12

then our bounds for the last integral is -y to y and our equation cancels out to 1/25

Answer 13

\[\int\limits_{-y}^{y}\frac{ 1 }{ 25 }\]

Answer 14

no. gotta be something like pi, rho, phi, etc.

Answer 15

@ganeshie8

Answer 16

dang sorry i think we just learned about rho today, wasn't paying attention in class though

Answer 17

looks like you're integrating over the solud region inside sphere of radius 5

Answer 18

whats the jacobian when u move from cartesian to spherical ?

Answer 19

he said he isn't gonna teach us jacobian. he said it in lecture today. we don't have to know it. but i know that x=rhosinphicostheta, etc

Answer 20

for bounds, try

\(\rho \) : 0->5
\(\large \theta\) : 0->2pi
\(\large \phi\) : -pi/2 -> pi/2

Answer 21

of (1/rho) * rho^2sinphi

Answer 22

okay, you just need to memorize the jacobian/volume scale factor also :
\[dxdydz = \rho^2\sin\phi d\rho d\theta d\phi \]

Answer 23

i'm getting zero for that.... http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=89c969c21b169fa996f899d9b2a98588&title=Spherical%20Integral%20Calculator&theme=blue&i0=(1%2Frho)rho%5E2sinphi&i1=drho%20dphi%20dtheta&i2=0&i3=5&i4=-pi%2F2&i5=pi%2F2&i6=0&i7=2pi&podSelect=&showAssumptions=1&showWarnings=1

Answer 24

try changing phi to 0->pi

Answer 25

i did. got 157.08 which wasn't right either.

Answer 26

make theta : -pi/2 --> pi/2

Answer 27

the link didn't work and when i changed theta, the widget says invalid

Answer 28

still not working. screen shot?

Answer 29

enter 25pi

Answer 30

ok. that worked.

Answer 31

that was annoying....

Answer 32

do u see why theta needs to be from -pi/2 to pi/2

Answer 33

i get where you got the 0 to 5 for rho from, but i cannot figure out when you're supposed to know if it's 0 to 2 pi or pi/2 or pi/4......

Answer 34

notice that x is from 0->5

Answer 35

\(\theta : -\pi/2 \to \pi/2\) :