Solving a surface integral

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##.

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##.

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]##

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).

Mr Davis 97