I am asked to find the area of surface $z=16-x^2-y^2$ on the first octant. I proceeded the following way:
Reasoning
$$A = \iint_D \left\vert \frac{\partial P}{\partial x} \times \frac{\partial P}{\partial y} \right\vert \ dA$$
Given $P = (x,y,16-x^2-y^2)$ we have
$$\frac{\partial P}{\partial x} = (1,0,-2x) \quad \frac{\partial P}{\partial y} = (0,1,-2y)$$
so that
$$\int_{0}^{\pi/2} \int_{0}^{4} r \sqrt{1+4r^2} \ drd\theta = \cdots = \frac{\pi}{24} \left( 65 \sqrt{65} -1 \right)$$
Is this correct?
Thank you.
Best Answer
Yes, it is correct. The same result can be obtained, by rotating around the $z$ axis, through an agle of $\frac{\pi}{2}$, the graph of $x(z)=\sqrt{16-z}$: $$S:=\frac{\pi}{2}\int_{0}^{16}x(z)\sqrt{1+(x'(z))^2}dz=\frac{\pi}{24} \left( 65 \sqrt{65} -1 \right).$$