Double and triple integrals to find volume below surface

Question

I want to find the volume in the first octant lying under the paraboloid $z=1-\frac{x^2}{4}-\frac{y^2}{81}$

I used the substitution $x=2r\cos\theta$ and $y=9r\sin\theta$ to obtain $$\iint (1-r^2)r drd\theta = \iint r-r^3 drd\theta$$.

But what are my bounds here? I assumed $0 \rightarrow 2\pi$ for the integral with respect to theta, but since we're in the first octant I'm thinking $\frac{\pi}{2}$ instead. Im not sure about the integral with respect to $r$ though, I thought from 0 to 2 first, but that doesn't seem to give me the right answer.

Answer 1

I see that there are some confusions in a different answer/comments. First a little bit about the geometry: the paraboloid goes in $z$ from $0$ to $1$, At every $z$ in this range, the projection one the $xy$ plane is an ellipse, with ratio of the axes $2:9$. At $z=0$ and $x=0$ we get maximum $y=9$, and $z=y=0$ implies maximum $x=2$.

Now we want to calculate the volume. We start with a volume element $\Delta z(x,y)dx dy$, where $\Delta z$ is the height from the $xy$ plane to the paraboloid. In this case $\Delta z(x,y)=z(x,y)=1-\frac{x^2}{4}-\frac{y^2}{81}$. The limits of integration are such as the $(x,y)$ point is in the quarter of the ellipse $1-\frac{x^2}{4}-\frac{y^2}{81}=0$, which we can write $x$ from $0$ to $2$ and $y$ from $0$ to $9\sqrt{1-\frac{x^2}{4}}$.

We can also write the same integral in a "cylindrical" or "polar" coordinate system. We do the transformation in the question $x=2r\cos\theta$ and $y=9r\sin\theta$. When I plug these in,the volume is going to be $$V=\iint 18(1-r^2) r drd\theta$$ The factor of $18$ appears from the Jacobian $J=18r$. The limits are $\theta=0$ to $\theta=\pi/2$ and $r=0$ to $r=1$

Answer 2

The inverse of OP's transformation \begin{cases} x &= 2r \cos \theta \\ y &= 9r \sin \theta \\ \rlap{x,y \ge 0} \end{cases} is \begin{cases} r &= \sqrt{\dfrac{x^2}{4}+\dfrac{y^2}{81}} \\ \theta &= \left\{ \begin{align} \arctan\left(\dfrac{y}{x}\right) &\quad \text{if } x > 0 \\ \frac\pi2 &\quad \text{if } x = 0 \end{align} \right. \end{cases} Since the solid lies in the first octant, we have $x,y,z \ge 0$. The paraboloid can be written as $z = 1 - r^2$. This suggests $0 \le r \le 1$ since $r$ is nonnegative.

The range of $\theta$ in the question ($[0,\frac\pi2]$) is correct.

The Jacobian is

$$\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right| = 18r$$

This gives the required volume

\begin{align} \iint_{\text{paraboloid}} (1-r^2)r drd\theta &= \int_0^{\pi/2} 18 \left(\int_0^1 (r-r^3) dr \right) d\theta \\ &= \int_0^{\pi/2} 18 \left( \frac12 - \frac14 \right) d\theta \\ &= 18 \cdot \frac14 \cdot \frac\pi2 \\ &= \frac{9\pi}{4} \end{align}