How would I find the integration limits for this volume integral?

Question

If i want to integrate over the the region $R$ such that $R:={(x,y,z):z\geq x^2+y^2,z \geq 2-2x, z\leq 10 -2y}$, what will the limits be for $x, y$, more importantly how should I systematically think to find the limits for my integrals? Could someone please help me? Is it correct if my integration limits are as such: $\int^{1}_{0} \int^{1}_{0} \int^{10-2y}_{2-2x} dzdxdy$?

Answer 1

For lack of being able to come up with a more straightforward approach (read: integrating over $R$ directly), you can try first finding the volume of the region between the upper plane $z=10-2y$ and the paraboloid $z=x^2+y^2$, then subtracting the volume of the region between the lower plane $z=2-2x$ and $z=x^2+y^2$.

$$\iiint_R\mathrm dV=\left\{\iiint_{R_1}-\iiint_{R_2}\right\}\mathrm dV$$

For either integral, I think converting to cylindrical coordinates is your best bet, but in each case you'll need a unique modification. For the first integral, let

$$\begin{cases}x=r\cos\theta-1\\y=r\sin\theta\\z=z\end{cases}\implies\mathrm dV=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz$$

Then $z=x^2+y^2=(r\cos\theta-1)^2+(r\sin\theta)^2=r^2-2r\cos\theta+1$ and $z=2-2x=2-2(r\cos\theta-1)=4-2r\cos\theta$, and this volume is

$$\iiint_{R_1}\mathrm dz\,\mathrm dy\,\mathrm dx=\int_0^{2\pi}\int_0^{\sqrt3}\int_{r^2-2r\cos\theta+1}^{4-2r\cos\theta}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\frac{9\pi}2$$

As the comment above points out, the plane $z=2-2x$ intersects the paraboloid in the cylinder $(x+1)^2+y^2=3$, i.e. with radius $\sqrt3$ and central axis parallel to the $z$ axis running through the "center" $(-1,0,0)$.

For the second integral, take

$$\begin{cases}x=r\cos\theta\\y=r\sin\theta-1\\z=z\end{cases}\implies\mathrm dV=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz$$

Then $z=x^2+y^2=(r\cos\theta)^2+(r\sin\theta-1)^2=r^2-2r\sin\theta+1$ and $z=10-2y=12-2r\sin\theta$. The volume is

$$\iiint_{R_2}\mathrm dV=\int_0^{2\pi}\int_0^{\sqrt{11}}\int_{r^2-2r\sin\theta+1}^{12-2r\sin\theta}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\frac{121\pi}2$$

Again, these limits come from the fact that the intersection between the plane $z=10-2y$ and the paraboloid occurs in the cylinder $x^2+(y+1)^2=11$, i.e. with radius $\sqrt{11}$ and axis parallel to the $z$ axis passing through the point $(0,-1,0)$.

Then the total volume of $R$ is

$$\iiint_R\mathrm dV=\frac{121\pi}2-\frac{9\pi}2=\boxed{56\pi}$$