Question

cylindrical coordinates:
find triple integral of : f(x,y,z) =5sqrt(x^2+y^2)
over volume x^2+y^2<4, 0 13 years ago

Answer 1

Converting to cylindrical, you have
\[\int\int\int_R5\sqrt{x^2+y^2}~dV~\iff~\int\int\int_R5r~(r~dr~d\theta~dz)\]
where \(R\) is the region defined by the following (in rectangular coordinates):
\[R:=\left\{(x,y,z):-2< x<2,~-2< y<2,~0<z<9-x-2y~\right\}\]
and in cylindrical coordinates:
\[R:=\left\{(r,\theta,z):0<r<2,~0<\theta<2\pi,~0<z<9-r\cos\theta-2r\sin\theta~\right\}\]
So the integral is now
\[5\int_0^2\int_0^{2\pi}\int_0^{{\large9-r\cos\theta-2r\sin\theta}}r^2~dz~d\theta~dr\]

Answer 2

I love it when i get the right answer!!! thank you for confirming!!!!

Answer 3

You're welcome! By the way, I'm getting \(240\pi\) as the final answer.

Answer 4

hmm got setup right must have wanked on the math somewhere i got 120pi but the setup is right so I can see where my mistake is.

Answer 5

I also could have made a mistake somewhere. I'll check my work again, but I'm confident in that integral setup.

Answer 6

Well, here's what I have. Everything seems to follow:
\[5\int_0^2\int_0^{2\pi}\int_0^{{\large9-r\cos\theta-2r\sin\theta}}r^2~dz~d\theta~dr\\
5\int_0^2\int_0^{2\pi}r^2\left(9-r\cos\theta-2r\sin\theta-0\right)~d\theta~dr\\
5\int_0^2\int_0^{2\pi}\left(9r^2-r^3\cos\theta-2r^3\sin\theta\right)~d\theta~dr\\
5\int_0^2\left(9r^2(2\pi-0)-r^3\left(\sin(2\pi)-\sin0\right)+2r^3\left(\cos(2\pi)-\cos0\right)\right)~dr\\
5\int_0^218\pi r^2~dr\\
90\pi\int_0^2r^2~dr\\
\frac{90}{3}\pi \left(2^3-0^3\right)\\
240\pi\]
Let me know if you see a mistake somewhere.