we have
$\iint_S z^2 dS$ , where $S$ is $x = y^2 + z^2$ for $0 \le x \le 1$
i have already simplified the integral using correct parameterizations and have
$1)$ $\iint_S z^2 dS$ $=$ $\iint_S z^2\sqrt{1+4(y^2 + z^2)}dS$
To further simplify the integral i will convert using polar coordinates $y=rcos\theta$ , $z=rsin\theta$ which simplifies to
$2)$ $\iint_D r^2sin^2\theta \sqrt{1 + 4r^2} r dA$
This is where i am stuck, i know everything up to $1)$ is correct. but it looks like converting to polar coordinates makes the integral more difficult. So my question is how to solve integral 1) thanks
When I first learnt how to do surface integrals it was always helpful to draw the surface.
If you struggle to visualize a surface then a good way to understand the shape is to draw it in 4 parts -> the surface in the x-y plane, with z=0 -> the surface in the x-z plane, with y=0 -> the surface in the y-z plane, with x=0 -> and finally the 3D surface in the x-y-z plane if it would help you understand the calculation
Normally you would take the equation you are given for the surface and rewrite it in the form z=f(x,y) because dA=dxdy
Then in the surface integral you substitute f(x,y) in place of z and the bounds are given by the values or graphs for x and y
In your case you are given the bounds for x, 0 and 1. Typically the bounds of either x or y will be given as constants and from there you set the bounds for the other variable in terms of the variable that you were given then constants for -> e.g. since you were given the bounds of x as constants, your bounds of y may be written as y=some function involving x, and then you integrate in terms of y first, then in terms of x.
After you’ve correctly substituted it becomes easier with practice to tell if converting to polar coordinates would be helpful or not.
I hope that this is maybe helpful somehow...