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Find the volume bounded by the elliptic paraboloids given by $z=x^2 + 9 y^2$ and $z= 18- x^2 - 9 y^2$.

First I found the intersection region, then I got $x^2+ 9 y^2 =1$. I think this will be area of integration now what will be the integrand. Please help me.

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  • $\begingroup$ I did upvote the answers but it is showing as I need 15votes for pubilcly display mu upvotes $\endgroup$ – Balaji Dec 2 '15 at 6:07
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You probably got something looking like this3D

where the required volume is the dark volume inside the two parabolas. Of course there are many ways to find that volume; I'll stick to the slice method: essentially you fix the third variable $z$ and calculate the area of desired region intersecting the plane $z = z_0$, then you sum up all this areas. Note that if you intersect the region with a plane parallel to the $XY$ plane you get an ellipse, with major semi-axis $\sqrt{\frac{18-z}{9}}$ and minor semi-axis $\sqrt{18-z}$ if $z \ge 9$ (that is when the parabolas intersect each other), and major semi-axis $\sqrt{z}$ and minor semi-axis $\sqrt{\frac{z}{9}}$ if $z \ge 9$ when $z \ge 9$. So the area of the intersection between the region and the plane at height $z$ is $$ \begin{cases} \sqrt{\frac{18-z}{9}} \cdot \sqrt{18-z} \cdot \pi = \frac{18 - z}{3} \pi \iff z \ge 9\\ \sqrt{\frac{z}{9}} \cdot \sqrt{z} \cdot \pi= \frac{z}{3} pi \iff z \le 9 \end{cases} $$ So the volume desired is $$\int \limits_0 ^{9} \frac{z}{3} \pi dz + \int \limits_9 ^{18} \frac{18 - z}{3} \pi dz = \pi \left(\frac13 \int \limits_0 ^9 z dz + \int \limits_9 ^{18} 6 dz - \frac13 \int \limits_9 ^{18} zdz \right) = \pi \left(\frac13 \frac{z^2}{2} \big|_0 ^9 + 6 \big|_9 ^{18} - \frac13 \frac{z^2}{2} \big|_9 ^{18} \right) = 27\pi$$

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  • $\begingroup$ Congratulations on the graphic in Geogebra. I also tried that but I am inexperienced in its 3D graphing. I couldn't adjust the graph just right. $\endgroup$ – Rory Daulton Dec 1 '15 at 13:02
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You are wrong about the intersection region. Equating the two expressions for $z$ gives

$$x^2+9y^2=18-(x^2+9y^2)$$ $$2(x^2+9y^2)=18$$ $$x^2+9y^2=9$$

and thus here $z=9$.

The volume of your intersection can be divided into to parts: $0\le z\le 9$ where the restrictions on $x$ and $y$ are $x^2+9y^2\le z$, and $9\le z\le 18$ where the restrictions on $x$ and $y$ are $18-(x^2+9y^2)\ge z$. You can see that those two parts have equal shapes and sizes and thus equal volumes, but that is not necessary to use.

So for each region, for each $z_0$ find the area of the cross-section of your region with the plane $z=z_0$, which is an ellipse so the area is easy to find and is an expression in $z_0$. Then integrate that area over $z$ between the limits I gave. Or if you like, use a triple integral for each region. It is also possible to do a double integral over the area $x^2+9y^2=9$.


Since you ask, I'll give more details. I prefer the single-integral approach, so I'll show that here.

For the lower region $x^2+9y^2\le z$ for $0\le z\le 9$, we can use our knowledge of conic sections to see that for a given $z$ that is an ellipse with major axis $a=\sqrt z$ over the $x$-coordinate and minor axis $b=\frac{\sqrt z}3$ over the $y$-coordinate. We can use the formula for the area of an ellipse

$$A=\pi ab=\pi(\sqrt z)\left(\frac{\sqrt z}3\right)=\frac{\pi z}3$$

We now find the volume of that region with

$$V_1=\int_0^9 \frac{\pi z}3\,dz$$

For the upper region $x^2+9y^2\le 18-z$ for $9\le z\le 18$, we can use our knowledge of conic sections to see that for a given $z$ that is an ellipse with major axis $a=\sqrt{18-z}$ over the $x$-coordinate and minor axis $b=\frac{\sqrt{18-z}}3$ over the $y$-coordinate. We can use the formula for the area of an ellipse

$$A=\pi ab=\pi(\sqrt{18-z})\left(\frac{\sqrt{18-z}}3\right)=\frac {\pi(18-z)}3$$

We now find the volume of that region with

$$V_2=\int_9^{18} \frac{\pi(18-z)}3\,dz$$

Your total volume is then $V_1+V_2$.

I like this approach since it is just a pair of single integrals, each of which is very easy. Your question seems to assume the double-integral approach. Let me know if those are the bounds you really want.


Here is the double-integral, if you really want it.

We saw that the largest possible area for a given $z$ is $x^2+9y^2\le 9$. We get from that

$$-1\le y\le 1, \qquad -3\sqrt{1-y^2}\le x\le 3\sqrt{1-y^2}$$

and the bounds on $z$ from your two original conditions are

$$x^2+9y^2\le z\le 18-x^2-9y^2$$

So the appropriate double integral is

$$\int_{-1}^1 \int_{-3\sqrt{1-y^2}}^{3\sqrt{1-y^2}} [(18-x^2-9y^2)-(x^2+9y^2)]\,dx\,dy$$

Good luck with that!

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  • $\begingroup$ please specify the limits of integration of x and y and the integrand explicitly. $\endgroup$ – Balaji Dec 1 '15 at 12:18
  • $\begingroup$ @Balaji: Is my addition good enough? Or do you really want to go the double-integral over $x$ and $y$ route, which is more difficult? $\endgroup$ – Rory Daulton Dec 1 '15 at 12:36
  • $\begingroup$ Your answer is very nice and easy. But I very curious to know the limits and integrand of double integration $\endgroup$ – Balaji Dec 1 '15 at 12:44

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