How Do You Choose the Correct Polar Coordinates for Surface Integrals?

[Mr Davis 97]

Homework Statement

Solve the surface integral $\displaystyle \iint_S z^2 \, dS$, where $S$ is the part of the paraboloid $x=y^2+z^2$ given by $0 \le x \le 1$.

Homework Equations

The Attempt at a Solution

First, we make the parametrization $x=u^2+v^2, \, y=u, \, z = v$, so let $\vec{r}(u,v) = \langle u^2 + v^2, u, v \rangle$, then through computation $| \vec{r}_u \times \vec{r}_v | = \sqrt{1+4u^2+4v^2}$, and so $\displaystyle \iint_S z^2 \, dS = \displaystyle \iint_D v^2 \sqrt{1+4u^2 + 4v^2} \, dA$, where $D = \{(u,v) \, | \, u^2 + v^2 \le 1 \}$. However, this is where I get stuck, because I want to use polar coordinates to evaluate the latter integral, but I am not sure whether to use $v = r \sin \theta$ or $v = r \cos \theta$

 

[member 587159]

Homework Statement

Solve the surface integral $\displaystyle \iint_S z^2 \, dS$, where $S$ is the part of the paraboloid $x=y^2+z^2$ given by $0 \le x \le 1$.

Homework Equations

The Attempt at a Solution

First, we make the parametrization $x=u^2+v^2, \, y=u, \, z = v$, so let $\vec{r}(u,v) = \langle u^2 + v^2, u, v \rangle$, then through computation $| \vec{r}_u \times \vec{r}_v | = \sqrt{1+4u^2+4v^2}$, and so $\displaystyle \iint_S z^2 \, dS = \displaystyle \iint_D v^2 \sqrt{1+4u^2 + 4v^2} \, dA$, where $D = \{(u,v) \, | \, u^2 + v^2 \le 1 \}$. However, this is where I get stuck, because I want to use polar coordinates to evaluate the latter integral, but I am not sure whether to use $v = r \sin \theta$ or $v = r \cos \theta$

It doesn't matter what you use. If you choose $v = r\cos(\theta)$, then you will have polar coordinates in the $zy$-plane. Thus, the angle starts from the $z$-axis. In the other case, in the $yz$-plane, the angle starts at the $y$ axis. The funny thing is that you don't even have to understand what's happening geometrically, as it is always a full rotation.

However, I should add, that there is a better parametrisation: (use cylindrical coordinates)

$\begin{cases}x = (R\cos(\theta))^2 + (R\sin(\theta))^2 = R^2 \\ y = R\cos(\theta) \\z=R\sin(\theta)\end{cases}$

where $R \in [0,1], \theta \in [0, 2\pi]$

 

[Mr Davis 97]

It doesn't matter what you use. If you choose $v = r\cos(\theta)$, then you will have polar coordinates in the $zy$-plane. Thus, the angle starts from the $z$-axis. In the other case, in the $yz$-plane, the angle starts at the $y$ axis. The funny thing is that you don't even have to understand what's happening geometrically, as it is always a full rotation.

However, I should add, that there is a better parametrisation: (use cylindrical coordinates)

$\begin{cases}x = (R\cos(\theta))^2 + (R\sin(\theta))^2 = R^2 \\ y = R\cos(\theta) \\z=R\sin(\theta)\end{cases}$

where $R \in [0,1], \theta \in [0, 2\pi]$

Would there every be a case when it matters which I choose? For example, what if I just had
$\displaystyle \iint_D v \, dA$, where $0 \le v^2 + u^2 \le 25$? Then wouldn't it matter?

 

[member 587159]

Would there every be a case when it matters which I choose? For example, what if I just had
$\displaystyle \iint_D v \, dA$, where $0 \le v^2 + u^2 \le 25$? Then wouldn't it matter?

It never matters if you can correctly interpret what's happening geometrically. In this case, it doesn't matter since $\theta \in [0,2\pi]$. If $\theta \in [0,a]$ with $a < 2\pi$, things get trickier.

Your question is easier to answer if you think just in the original 2D $xy$ plane (that's how I think about it. If we use polar coordinates in yz, then just act as the the yz plane is the xy plane and change y,z accordingly).