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Find the volume between $z=x^2$ and $z=4-x^2-y^2$

I made the plot and it looks like this:

enter image description here

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?

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2 Answers 2

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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

$$\begin{cases} x & = \frac{r \cos \theta}{\sqrt{2}} \\ y & = r \sin \theta, \end{cases}$$

and the "area element" becomes

$$dA = \frac{r}{\sqrt{2}} \, dr \, d \theta.$$

This is obtained via the Jacobian. I can explicitly write it out if you need. Proceeding the computations yields

$$ \begin{align} \iint 4 -2x^2 -y^2 \, dA & = \int_0^{2 \pi} \hspace{-5pt} \int_0^2 (4-r^2) \frac{r}{\sqrt{2}} \, dr \, d \theta \\ & = \frac{2 \pi}{\sqrt{2}} \int_0^2 4r - r^3 \, dr \\ & = \frac{2 \pi}{\sqrt{2}} \left( 2r^2 \bigg\vert_0^2 - \frac{r^4}{4} \bigg\vert_0^2 \right) \\ & = \frac{2 \pi}{\sqrt{2}} \left( 8 - \frac{16}{4} \right) \\ & = \frac{2 \pi}{\sqrt{2}} \cdot \frac{16}{4} \\ & = \frac{32 \pi}{4 \sqrt{2}} = 4\sqrt{2}\pi. \end{align} $$

Best wishes. :)

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I agree with Fantini that you do not need the $z$ inside the integral. Generally, when you see a lot of squares, especially $x^2+y^2$, then you should be thinking about switching to polar/cylindrical coordinates. This will get the answer for this integral, though since the boundary is an ellipse it is not quite so straight-forward an integral. The equation of the ellipse becomes

$2r^2\cos^2\theta+r^2\sin^2\theta=4 \Longrightarrow r^2 = \dfrac{4}{\sin^2\theta+2\cos^2\theta} = \dfrac{4}{1+\cos^2\theta}$

So I get the following result:

$\begin{align} \iint 4 -2x^2 -y^2 \, dA &= \iint 4-(x^2+y^2) - x^2 dA \\ & = 4\int_0^{\pi/2} \hspace{-5pt} \int_0^{\sqrt{\frac{4}{1+\cos^2\theta}}} (4-r^2-r^2\cos^2\theta) \, r \, dr \, d \theta \\ &= 4\int_0^{\pi/2} \hspace{-5pt} \int_0^{\sqrt{\frac{4}{1+\cos^2\theta}}} (4r-r^3-r^3\cos^2\theta) \, dr \, d \theta \\ &= 4\int_0^{\pi/2} \hspace{-5pt} (2r^2-\frac{1}{4}r^4-\frac{1}{4}r^4\cos^2\theta)\bigg\vert_0^{\sqrt{\frac{4}{1+\cos^2\theta}}} \, d \theta \\ & = 4\int_0^{\pi/2} \dfrac{8}{1+\cos^2\theta} - \dfrac{4}{(1+\cos^2\theta)^2} - \dfrac{4\cos^2\theta}{(1+\cos^2\theta)^2} \, d \theta \\ & = 4\int_0^{\pi/2} \dfrac{8+8\cos^2\theta}{(1+\cos^2\theta)^2} - \dfrac{4+4\cos^2\theta}{(1+\cos^2\theta)^2} \, d \theta \\ & = \int_0^{\pi/2} \dfrac{16+16\cos^2\theta}{(1+\cos^2\theta)^2} \, d \theta \\ & = \int_0^{\pi/2} \dfrac{16}{1+\cos^2\theta} \, d \theta \\ &= \int_0^{\pi/2} \dfrac{16}{\sin^2\theta+2\cos^2\theta} \, d \theta\\ & = \int_0^{\pi/2} \dfrac{16}{\sin^2\theta+2\cos^2\theta}\cdot\dfrac{\frac{1}{\cos^2\theta}}{\frac{1}{\cos^2\theta}} \, d \theta \\ &= \int_0^{\pi/2} \dfrac{16\sec^2\theta}{\tan^2\theta+2} \, d \theta \\ &= \int_0^{\infty} \dfrac{16}{u^2+(\sqrt{2})^2} \, d u \\ &= \dfrac{16}{\sqrt{2}} \tan^{-1}\left(\dfrac{u}{\sqrt{2}}\right) \bigg\vert_0^{\infty} = \dfrac{16}{\sqrt{2}}\cdot\dfrac{\pi}{2} = 4\sqrt{2}\pi \end{align}$

Note that I've done a bit of hand-waving at the end to deal with the improper integral and that the above solutions agrees with a computer calculation of the answer.

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  • $\begingroup$ Not all that much "hand-waving": Dante mentions Stewart, so they would have covered that particular improper integral well before discussing a volume integration of this sort. $\endgroup$ Jan 25, 2014 at 3:44

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