Question

Find double integral volume of a solid in first octant bounded by coordinate plane, the cylinder x^2+y^2=4, and z+y=3

Answer 1

Cylindrical coordinate conversion is your friend here. If \(R\) is the region of interest, then you can describe it by
\[R:\left\{(x,y,z)~:~0\le x\le2,~0\le y\le2,~0\le z\le3-y\right\}\]
in rectangular coordinates, and
\[R:\left\{(r,\theta,z)~:~0\le r\le2,~0\le\theta\le\frac{\pi}{2},~0\le z\le3-r\sin\theta\right\}\]
in cylindrical. Does that make sense?

Answer 2

The volume would then be given by the integral,
\[V=\int\int\int_RdV=\int_0^{\pi/2}\int_0^2\int_0^{3-r\sin\theta}r~dz~dr~d\theta\]

Answer 3

Oh wait, you need a double integral representation... Hmm...

Answer 4

Thank you so much for the reply.
So would V= integral from 0 to 2 integral from 0 to 2 of 3-y dx dy?

Answer 5

No, you have to convert to polar/cylindrical for this one (double for polar, triple for cylindrical).

My definition of \(R\) has a mistake, actually. It should be
\[R:\left\{(x,y,z)~:~0\le x\le2,~0\le y\le\color{red}{\sqrt{4-x^2}},~0\le z\le3-y\right\}\]
The red is what's preventing you from using rectangular coordinates.

Here's the double integral expression:
\[V=\int\int_RdV=\int_0^{\pi/2}\int_0^2(3-r\sin\theta)~r~dr~d\theta\]
(which is actually equivalent to the triple integral I posted earlier.

Answer 6

Thank you :)

Answer 7

yw