Value of surface Integral Let S be the part of the surface $z = x^2+y^2$ which lies under the plane $z=4$.
What is the value of the surface integral $\iint_S z \, ds \, \,$?
What I've done so far:
I have used the surface integral formula however I do not know if I need to convert to polar coordinates for the boundaries when integrating? i can't seem to figure out the boundaries.
I have set up the interval as follows:
$\int \int (x^2+y^2) \sqrt{1+4(x^2+y^2)} \, dx \, dy$

$\displaystyle \iint_S z \, dS = \iint (x^2+y^2) \sqrt{1+4(x^2+y^2)} \, \,dx \,dy \, \,$ is correct.
$x^2 + y^2 = r^2 = z \leq 4 \,$ so $ \,0 \leq r \leq 2$. Your integral becomes
$\displaystyle  =\int_0^{2\pi} \int_0^2 r^2 \sqrt{1+4r^2} \, \, r \, dr \, d \theta = \int_0^{2\pi} \int_0^2 r^3 \sqrt{1+4r^2} \, dr \, d \theta$
Can you take it from here?
