Hello,
Did you multiply by the determinant of the Jacobian matrix when changing to cylindrical coordinates ?
Hi,
So my task is to evaluate this triple integral:
SSS yz DV, over the region E: above z = 0, below z = y, and inside the cylinder x^2 + y^2 = 4.
I thought it'd be best to use cylindrical coordinates. Thus:
0 <= z <= rsinθ
0 <= r <= 2
0 <= θ <= 2π
But I keep getting 0. The answer is actually 64/15.
Here are the steps I took to compute the integral:
= SSS r^2 * sinθ * z dz dr dθ
= SS r^2 * sinθ * [(z^2)/2] from 0 to rsinθ .. dr dθ
= SS (r^4 * (sinθ)^3)/2 dr dθ
= S ((sinθ)^3)/2 * [(r^5)/5] from 0 to 2 dθ
= (16/5) S ((sinθ)^3) dθ
use the trig identity to split it up (sinθ)^2 = 1 - (cosθ)^2, the use substitution... etc
the final result:
(16/5) * [((cosθ)^3)/3 - cosθ] from 0 to 2π
This is obviously zero, since cos(0) and cos(2π) produce the same result. What am I doing wrong?
y = r sinθ
z = z
Oh, whoops.. I guess I did multiply it by the determinant of the Jacobian matrix... I really should start looking up the proper definitions of things... I didn't exactly know the name of it... and of course, this is just a plug-and-chug calculus course... so the lecturers dont bother to explain anything rather than just say "you must do $BLAH to do this"
So...
rsinθ * z * r * dz dr dθ
Solving it with Cartesian coordinates works... but I don't understand why doing it in polar does not... is it because theta should be between 0 and pi and not 0 and 2pi?
EDIT: Yep! That's. D'oh! What a stupid mistake... It's because z is bounded by 0 and y... and if z's lower bound is 0, you obviously can only have the top half of the circle in the xy-plane... =) which is from 0 to pi