Recent content by ishanz

Got it. Thanks very much, Dick.

I see, that makes sense. Are there any situations in which I would ever have to actually worry about a Jacobian factor when doing a surface integral similar to the one I've described above? Or is it something I should only worry about when explicitly executing coordinate transforms?

God, I'm bad at this whole Latex + forum thing. I'm so sorry about the double posts... I think I accidentally hit submit or something. I don't quite understand. I don't know if it's because of the totally methodical and unintuitive way that our professor has taught us (i.e., simply equating...

I don't quite understand. I don't know if it's because of the totally methodical and unintuitive way that our professor has taught us (i.e., simply equating dA to rdrd\theta in all scenarios) or because of my own negligence. Does the |{\bf R}_r\times{\bf R}_\theta| factor always take care of...

Sorry, the final step of the integration process didn't come out right. Here it is: \int_0^{2\pi}\int_1 ^3 {(rcos(\theta))^2(r)^2(r\sqrt{2})}drd\theta=\frac{364\sqrt{2}\pi}{3}

Homework Statement Evaluate the surface integral. ∫∫S x^2*z^2 dS S is the part of the cone z^2 = x^2 + y^2 that lies between the planes z = 1 and z = 3. Homework Equations \int \int _{S}F dS = \int \int _D F(r(u,v))|r_u\times r_v|dA x=rcos(\theta) y=rsin(\theta) The Attempt...

Ahh, that makes a lot of sense. Thanks a lot!

Homework Statement Convert t = pi/6 into polar form. Homework Equations x = r*cos(t) y = r*sin(t) The Attempt at a Solution t = pi/6 cos(t) = cos(pi/6) cos(t) = sqrt(3)/2 x/r = sqrt(3)/2 sqrt(3)r = 2x sqrt(3)sqrt(x^2 + y^2) = 2x sqrt(3x^2 + 3y^2) = 2x 3x^2 + 3y^2 = 4x^2...