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**Reskal** Problem: Find the surface area of the paraboloid $\displaystyle z=4-x^2-y^2$ that lies above the xy-plane. Finding where the function crosses the xy plane- $\displaystyle 4=x^2+y^2$ taking the gradient- $\displaystyle \{2x,2y\}$ Useing the surface area formula-

$\displaystyle \int\int\sqrt{(4x^2+4y^2+1)}dxdy$

This is where I'm unsure of how to continue. It seems that I could just set my bounds to be [-2,2] and [0,2] but this gives and incorrect answer. [That is because you would then be calculating the surface area over the square $\displaystyle \color{red}-2\leqslant x\leqslant2,\ -2\leqslant y\leqslant2$.]

The solution book says to convert the function to polar coordinates as such -

$\displaystyle \int_{0}^{2\pi}\int_{0}^{2}\sqrt{(4r^2+1)}drd\thet a$, [Correct except that $\displaystyle \color{red}drd\theta$ should be $\displaystyle \color{red}r\,drd\theta$.]

I'm unsure of the bounds for theta, wouldn't 0-2\pi give the surface area of the whole sphere [It's not a sphere, it's a paraboloid, and you're calculating the area of that part of its surface that lies above the disk $\displaystyle \color{red}x^2+y^2\leqslant4$, which in polar coordinates corresponds to $\displaystyle \color{red}0\leqslant\theta\leqslant2\pi,\ 0\leqslant r\leqslant2$.] and not the area above $\displaystyle xy$? How would you compute the total surface area if this equation just gives it to you above the xy-plane?