Compute the surface of $M=\{(x,y,z):2z=x^2+y^2,z<2\}$

Question

Compute the surface of $M=\{(x,y,z):2z=x^2+y^2,z<2\}$.

EDIT

I've deleted my previous attempts, because they were wrong (as noted in the comments). While I have accepted an answer, I've written down my own approach here below.

First we find a parametrization for our paraboloid. For $r\in(0,2),\theta\in(0,2\pi)$ we define $$ \phi(r,\theta)=(r\cos\theta,r\sin\theta,r^2/2). $$ The image of $\phi$ yields $M$, except for a set of measure zero, so that is ok. We want to compute the following integral: $$ \int_MdV=\int_{(0,2)\times(0,2\pi)}V(D\phi). $$ So we need the volume element $V(D\phi)=\sqrt{\operatorname{det}(D\phi^TD\phi)}$. We have $$ D\phi(r,\theta)=\begin{pmatrix} \cos\theta&-r\sin\theta\\ \sin\theta&r\cos\theta\\ r&0 \end{pmatrix}, $$ which yields $$ D\phi^TD\phi=\begin{pmatrix}1+r^2&0\\0&r^2\end{pmatrix}. $$ So we have $V(D\phi)=\sqrt{r^2(1+r^2)}=r\sqrt{1+r^2}$ (we can drop the absolute value for $r$, because $r>0$). At last, we compute the 2-dimensional volume: $$ \int_{r=0}^2\int_{\theta=0}^{2\pi}r\sqrt{1+r^2}\,d\theta dr=2\pi\int_{r=0}^2r\sqrt{1+r^2}\,dr=\dfrac{2\pi}{3}(5\sqrt 5-1).$$

Answer 1

In polar coordinates we have

• $0\le z \le 2$
• $0\le \theta \le 2\pi$
• $r(z)=\sqrt{2z}$

also note that

• $\frac{dr}{dz}=\frac1{\sqrt{2z}}\implies dr=\frac1{\sqrt{2z}}dz\implies ds=\sqrt{dz^2+dr^2}=\sqrt{\frac{1+2z}{2z}}dz$

then

$$S=\int_0^{2\pi}d\theta\int_0^2 r(z)\,ds=2\pi\int_0^2 \sqrt{2z}\sqrt{\frac{1+2z}{2z}}\,dz=2\pi\int_0^2 \sqrt{1+2z}\,dz$$

As an alternative by your parametrization the surface integral is

$$S=\iint_M|\vec \phi_r\times \vec \phi_\theta|\,dr\,d\theta$$

with

• $\vec \phi_r=(\cos \theta,\sin \theta,r)$
• $\vec \phi_\theta=(-r\sin \theta,r\cos \theta,0)$

and then

• $\vec \phi_r\times \vec \phi_\theta=(-r^2\cos \theta,-r^2\sin \theta,r)$
• $|\vec \phi_r\times \vec \phi_\theta|=\sqrt{r^4+r^2}=r\sqrt{1+r^2}$

and finally

$$S=\iint_Mr\sqrt{1+r^2}\,dr\,d\theta=\int_0^{2\pi} d\theta \int_0^2 r\sqrt{1+r^2}\,dr=2\pi\left[\frac13(1+r^2)^\frac32\right]_0^2=\frac{2\pi}3\left(5\sqrt 5 -1\right)$$

Answer 2

Noe That the surface area of $$M=\{(x,y,z):2z=x^2+y^2,z<2\}$$ is the surface of revolution generated by rotating the region in $xz$ plane bounded by $z=\frac {1}{2} x^2$, $x=0$ , and $z=2$ about the $z-axis.$

Thus the surface area is $$ S=\int _0^2 2\pi x ds = \int _0^2 2\pi x \sqrt {1+z'^2}dx=\\ \int _0^2 2\pi x \sqrt {1+x^2}dx =\\ \frac {2\pi}{3}(5\sqrt 5 -1) $$