Use cylindrical or spherical coordinates, whichever seem more appropriate,
to evaluate the triple integral $\displaystyle \int \int \int_{V} z dV $ where V lies above the paraboloid $\displaystyle z = x^2 + y^2$ and below the plane $\displaystyle z = 6y.$
I am having trouble with the bounds and am not sure whether to use cylindrical or spherical coordinates.
I guess I can see it in cylindrical:
z is going...
from r^2 (the paraboloid) up to 6y i.e. 6r sin theta (the plane roof)
... for every bit that r is going...
from zero (the z axis) along to as far as where r^2 = 6r sin theta => r = 6 sin theta
... for every bit that theta is going from zero (the +ive x axis) round to pi (-ive). So,
$\displaystyle \displaystyle{V = \int_{0}^{pi}\ \int_{0}^{6 \sin \theta} \int_{r^2}^{6r \sin \theta} rz\ dz\ dr\ d\theta}$
int from 0 to pi int from 0 to 6sin theta int from r^2 to 6rsin theta rz dz dr dtheta - Wolfram|Alpha
The integral graph there (going 0 to pi on the horizontal) looks about right.
Those zero sins aren't going to work, but see also,
integrate 1944 sin^6 theta - Wolfram|Alpha
which gives 121.5 pi. If this isn't wrong I'll put a pic for the computation similar to
http://www.mathhelpforum.com/math-he...tml#post621761
when I'm home, could be a while.
Anyway, hope that helps.
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
  |
LinkBack URL
About LinkBacks
