calculusintegrationmultivariable-calculus
I have to find the volume below the plane $z=3-2y$ and above the paraboloid $z=x^2+y^2$.
Integrating by $z$ first, it looks like the "arrow" I draw parallel to z-axis enters the region at $z=x^2+y^2$ and exits the region at $z=3-2y$. So $\int_{x^2+y^2}^{3-2y}1 dz$.
I'm confused how I am supposed to find the other two integrals. Any hints please?
$y = 2x^2 +2z^2$ and the plane $y=8$ gives:
$8=2x^2 +2z^2$
$4=x^2 +z^2$
This is a circle and gives you limits for $x$ and $z$ obviously it will be polar coordinates problems in xz space.
Limits for y are solved because your surface is going from paraboloid to the $y=8$, so modify your integral bounds... to be something like: $x=rcos(\phi), z=rsin(\phi)$, $y=y$:
$$\int_{0}^{2\pi}d\phi\int_{0}^{2}dr\int_{2r^2}^8 rdy$$
projecting the section of paraboloid and $y=8$ you get xz projection which defines you x and z bounds.
You found the intersection correctly. Here is a 2D cross section of how it would look when seen from the side on $YZ$ plane. It is bound below by the paraboloid $z = x^2+y^2$ and above by the plane $z = 2y$,
From $z = x^2 + y^2, z = 2y,$ you get $x^2 = 2y-y^2 \implies x = \pm \sqrt{2y-y^2}$
So your integral becomes,
$\displaystyle \int_{0}^{2}\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}(2y - x^2 - y^2) dx \, dy$
In cylindrical coordinates,
$x = r \cos \theta, y = r \sin \theta$
So $z = x^2 + y^2 =r ^2 = 2y = 2r \sin \theta \implies r = 2 \sin \theta$
So your integral becomes,
$\displaystyle \int_{0}^{\pi}\int_{0}^{2\sin\theta}(2r\sin \theta - r^2) \, r \, dr \, d\theta$
Best Answer
Set $x^2 + y^2 = 3 - 2y$ to find the intersection in the xy-plane. You can then get the limits for x by solving for it in the above equation, or you can complete the square to get $x^2 + (y + 1)^2 = 4$ and solve for y.