Find the volume of the solid above the $xy$ plane and directly below the portion of the elliptic paraboloid $x^2+\frac{y^2}{4}=z$ which is cut off by the plane $z=9$.
Solution 1:
$$ \begin{align} V &= \int\int(z_2-z_1)dxdy \\
&= \int\int{(9-x^2-\frac{y^2}{4})}dxdy \\
&= \int_0^3\int_0^{2\sqrt{9-x^2}}{(9-x^2-\frac{y^2}{4})}dxdy \end{align} $$
Solving this I got $ V=\frac{81\pi}{4}$
Solution 2:
I take strip of ellipse at some z as
$\frac{x^2}{z}+\frac{y^2}{4z}=1$
The area of this ellipse, $A = 2\pi z$
Hence
$$ \begin{align} V &= \int_{z=0}^{z=9}Adz \\
&= \int_0^92\pi zdz = 81\pi \end{align} $$
Which one is correct?
Best Answer
The issue is with the first solution. Why did you choose the lower limit to be $0$ for both $x$ and $y$? The outer integral should go from $-3$, the inner integral from $-2\sqrt{9-x^2}$ to $2\sqrt{9-x^2}$. Since each of the integrands is symmetric, you just need to multiply each by a factor of $2$, so a total of $4$. Then you get the same answer.