Re: what is the mistake here

Your prof is correct; you need to substitute spherical coordinates into the equation for the volume. Your limits are wrong because you are not integrating over a sphere of radius 2 centered at the origin.

Re: what is the mistake here

correct i am integration on sphere centered (0,0,1)

in spherecal coordinated the third coordinate is the radius and from the photo and you said it youself that the radius is 2.

when i change the coordinated from cartesian to spherecal

i do it from the sketch,so i get radius from 0 till 2

cant see how the shifted center changes the intervals the way it did.

Re: what is the mistake here

The spherical coordinates are with respect to the Cartesian axes, not the center of the sphere that you're integrating over. Substituting the expressions for the spherical coordinates into $\displaystyle x^2+y^2+z^2\le 2z$ gives $\displaystyle \rho\le 2\cos \phi$.

You can shift the Cartesian coordinates to the center of the sphere but then you need to make the corresponding change to the integrand.

Re: what is the mistake here

Quote:

Originally Posted by

**ojones** The spherical coordinates are with respect to the Cartesian axes, not the center of the sphere that you're integrating over. Substituting the expressions for the spherical coordinates into $\displaystyle x^2+y^2+z^2\le 2z$ gives $\displaystyle \rho\le 2\cos \phi$.

You can shift the Cartesian coordinates to the center of the sphere but then you need to make the corresponding change to the integrand.

ok ,but how mathematicky can i get this transition

how to get this expression

$\displaystyle \rho\le 2\cos \phi$

Re: what is the mistake here

You did something similar already on your exam; how else did you transform the integrand to $\displaystyle 1/\rho$?

Re: what is the mistake here

Quote:

Originally Posted by

**ojones** You did something similar already on your exam; how else did you transform the integrand to $\displaystyle 1/\rho$?

why would i want

$\displaystyle 1/\rho$

?

Re: what is the mistake here

My point is how did you get $\displaystyle 1/\rho$?

Re: what is the mistake here

Quote:

Originally Posted by

**ojones** My point is how did you get $\displaystyle 1/\rho$?

$\displaystyle \rho^2=x^2+y^2+z^2$

so i got it from putting the $\displaystyle \rho$ instead of the expression in the denominator

cant see how it changes the interval into this cosine on top

?

Re: what is the mistake here

You skipped a big step here. How is this identity obtained?

Re: what is the mistake here

i have some linking formulas

between spherecal and cartesian

but we dont use them

we just see the sketch and build the integral by it.

can you tell me from the transision formulas

why $\displaystyle \rho$

have these intervals?

Re: what is the mistake here

You don't use the expressions relating spherical to cartesian coordinates?? That's absurd. This is why you don't understand your professor's comments. Go back to the beginning of this subject and learn it properly this time. There are no shortcuts, believe me.

Re: what is the mistake here

i have some linking formulas

between spherecal and cartesian

but we dont use them

we just see the sketch and build the integral by it.

can you tell me from the transision formulas

why $\displaystyle \pho$

have these intervals?

you dont use these formulas to build the integral

you do it by the sketch

but here i dont know how to get the 2cosine interval

could you show me the transition?

Re: what is the mistake here

I'll give you a hint. What is $\displaystyle z$ in spherical coordinates?

Re: what is the mistake here

$\displaystyle z=\rho cos \theta$

what to do now?