Question: Find the volume of the solid bounded by $z=x^2+y^2+1$ and $z=2-x^2-y^2$.
Setting the 2 equations equal w.r.t. $z$, $x^2+y^2+1=2-x^2-y^2 \rightarrow x=\pm\sqrt{\frac 12-y^2}$
Therefore the boundary of $y=\pm\frac {1}{\sqrt2}$.
So to find the volume of the solid, take the integration by subtracting the volume above and below the boundaries.
$\displaystyle V=\int_{-\frac {1}{\sqrt2}}^{+\frac {1}{\sqrt2}}\int_{-\sqrt{\frac 12-y^2}}^{+\sqrt{\frac 12-y^2}}(2-x^2-y^2)dxdy-\int_{-\frac {1}{\sqrt2}}^{+\frac {1}{\sqrt2}}\int_{-\sqrt{\frac 12-y^2}}^{+\sqrt{\frac 12-y^2}}(x^2+y^2+1)dxdy$
This is what I did. Without solving the equation, can someone tell me if it is correct?
Thank you!
Best Answer
Your setup is right. Here is the method you could have done to compute the volume.
Assume the density is $f(x,y,z) = 1$, so
$$V = \iiint_D \,dx\,dy\,dz$$
We are given that the solid is bounded by $z = x^2 + y^2 + 1$ and $z = 2 - x^2 - y^2$. As I commented under your question, you need to use cylindrical coordinates to evaluate the volume integral. Using the substitutions $x = r\cos(\theta)$, $y = r\sin(\theta)$ and $z = z$, we have $z = r^2 + 1$ and $z = 2 - r^2$. With some knowledge in graphs and functions, we see that the bounds are
$$\begin{aligned} r^2 + 1 \leq z \leq 2 - r^2\\ 0 \leq \theta \leq 2\pi\\ 0 \leq r \leq \dfrac{1}{\sqrt{2}} \end{aligned}$$
where $r = \frac{1}{\sqrt{2}}$ is found by solving for $r$ when $r^2 + 1 = 2 - r^2$. So for the volume triple integral, we have
$$\begin{aligned} V &= \iiint_D \,dx\,dy\,dz\\ &= \iiint_D r\,dr\,d\theta\,dz\\ &= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}}\int_{r^2 + 1}^{2 - r^2}r\,dz\,dr\,d\theta\\ &= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} r(2 - r^2 - r^2 - 1)\,dr\,d\theta\\ &= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} r(1 - 2r^2)\,dr\,d\theta\\ &= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} (r - 2r^3)\,dr\,d\theta\\ &= \int_{0}^{2\pi}\,d\theta\left.\left(\dfrac{1}{2}r^2 - \dfrac{1}{2}r^4 \right)\right\vert_{r = 0}^{r = \frac{1}{\sqrt{2}}}\\ &= 2\pi \left.\left(\dfrac{1}{2}r^2 - \dfrac{1}{2}r^4 \right)\right\vert_{r = 0}^{r = \frac{1}{\sqrt{2}}}\\ &= 2\pi \cdot \dfrac{1}{8}\\ &= \dfrac{\pi}{4} \end{aligned}$$