find the volume of the region bounded by xy-plane the paraboloid $z = x^2+y^2$ and the elliptic cylinder $\frac{x^2}{9}+\frac{y^2}{4} = 1$
lets use elliptical coordinates:
$$x=3rcos\theta $$
$$y = 2rsin\theta$$
Let's find the Jacobian:
$$\frac{\partial (x,y)}{\partial (r,\theta)} = 6r $$
now, my question is what is the function of the volume bounded by the paraboloid and cylinder should we subtract both functions?
$$\int^? \int^? (?)6rdrd\theta$$
also I don't know how to find the bounds
Best Answer
Let $V$ be the volume of the region and $E$ be the intersection of the elliptic cylinder with the $xy$-plane. Then $V=\int\int_{E} z dE$. We parametrize $E$ by elliptical coordinates $(r,\theta)\in F$, where $F=[0,1]\times [0,2\pi]$ putting $x=3r\cos\theta $ and $y = 2r\sin\theta$. Then $\int\int_{E} z dE=\int\int_{F} z |J(r,\theta)|dF$, where $J(r,\theta)=\frac{\partial (x,y)}{\partial (r,\theta)} = 6r$. Thus $$V=\int\int_{F} z |J(r,\theta)|dF=\int_{0}^1 dr \int_{0}^{2\pi} (9r^2\cos^2\theta+4r^2\sin^2\theta)\cdot 6r d\theta=$$ $$\int_{0}^1 dr \left(54r^3\int_{0}^{2\pi} \cos^2\theta d\theta+24r^3\int_{0}^{2\pi}\sin^2\theta d\theta\right)=$$ $$\int_{0}^1 dr \left(54r^3\pi +24r^3\pi\right)= \int_{0}^1 78r^3\pi dr=\frac{39\pi}2.$$