Find the volume between $z=x^2$ and $z=4-x^2-y^2$
I made the plot and it looks like this:
It seems that the projection over the $xy$-plane is an ellipse, because if $z=x^2$ and $z=4-x^2-y^2$ then $2x^2+y^2=4$ which means that $\displaystyle\frac{x^2}{(\sqrt{2})^2}+\displaystyle\frac{y^2}{2^2}=1$
Stewart define a region I if it is of the kind $\{(x,y,z):(x,y)\in D, u_1(x,y)\leq z \leq u_2(x,y)\}$.
I believe that the region I'm asked for cab be describe setting $D=\{(x,y): \displaystyle\frac{x^2}{(\sqrt{2})^2}+\displaystyle\frac{y^2}{2^2}=1\}$ and then $E=\{(x,y,z):(x,y)\in D, x^2\leq z \leq 4-x^2-y^2\}$.
Can the volume $V(E)$ be computed by $\displaystyle\int\displaystyle\int_D\displaystyle\int_{x^2}^{4-x^2-y^2}z\;dz$?
Or maybe considering that $0\leq \sqrt{x} \leq \sqrt{2}$ and $0 \leq y \leq \sqrt{4-x^2-y^2}$ could I compute the volume calculating $\displaystyle\int_0^{\sqrt{2}}\displaystyle\int_0^{\sqrt{4-2x^2}}\displaystyle\int_{x^2}^{4-x^2-y^2}z\;dzdydx ?$
The issue with the approach above is that doesn't seem too easy after the first two integrals, because:
$\displaystyle\int_0^{\sqrt{2}}\displaystyle\int_0^{\sqrt{4-2x^2}}\displaystyle\int_{x^2}^{4-x^2-y^2}z\;dz = \displaystyle\frac{1}{2} \displaystyle\int_0^{\sqrt{2}}\displaystyle\int_0^{\sqrt{4-2x^2}} [(4-x^2-y^2)^2-x^2]\; dydz \\
=\displaystyle\frac{1}{2} \displaystyle\int_0^{\sqrt{2}} \left(16y-7x^2y-\frac{8}{3}y^3+\frac{2}{3}x^2y^3+x^4y+\frac{1}{5}y^5\right)\rvert_0^{\sqrt{4-2x^2}}\;dx$ which after the substitution seems hard to evaluate.
Is there an easy way?
Best Answer
The mistake in your solution is that the volume of the set $E$ isn't
$$\iiint\limits_{E} z \, dV, \text{ but } \iiint\limits_{E} \, dV.$$
So we have
$$V(E) = \iint \hspace{-5pt} \int_{x^2}^{4-x^2-y^2} \, dz \, dA = \iint 4-2x^2 -y^2 \, dA.$$
This can be tackled in a couple of ways. One is using cartesian coordinates, as you have. The other is using polar coordinates. Since $2x^2 +y^2 = 4$ we have
$$\frac{x^2}{(\sqrt{2})^2} + \frac{y^2}{2^2} = 1.$$
To use polar coordinates here we switch using
$$\begin{cases} x & = \frac{r \cos \theta}{\sqrt{2}} \\ y & = \frac{r \sin \theta}{2}, \end{cases}$$
and the "area element" becomes
$$dA = \frac{r}{2 \sqrt{2}} \, dr \, d \theta.$$
This is obtained via the Jacobian. I can explicitly write it out if you need. Proceeding the computations yields
$$ \begin{align} \iint 4 -2x^2 -y^2 \, dA & = \int_0^{2 \pi} \hspace{-5pt} \int_0^1 (4-r^2) \frac{r}{2\sqrt{2}} \, dr \, d \theta \\ & = \frac{2 \pi}{2 \sqrt{2}} \int_0^1 4r - r^3 \, dr \\ & = \frac{\pi}{\sqrt{2}} \left( 2r^2 \bigg\vert_0^1 - \frac{r^4}{4} \bigg\vert_0^1 \right) \\ & = \frac{\pi}{\sqrt{2}} \left( 2 - \frac{1}{4} \right) \\ & = \frac{\pi}{\sqrt{2}} \cdot \frac{7}{4} \\ & = \frac{7 \pi}{4 \sqrt{2}}. \end{align} $$
Best wishes. :)