Finding the surface area of the paraboloid $z = 1 - x^2 - y^2$ that lies above the plane $z = -4$

Question

Firstly, I converted the surface area integral into polar coordinates.

After simplifying the integral I integrated with respect to the radius:

Finally, I integrated with respect to our angle. However this answer is wrong, I have a feeling it might be my bounds. Since we are only considering the region from $z = 1$ to $z = -4$ that means between those two planes, the radius of the paraboloid is changing from $0$ to $5$, and since the cross-section are circles, our angle goes from $0$ to $2 \pi$. Is my reasoning incorrect? Or did I get something else wrong?

thanks

Answer 1

The surface can be viewed as the stack of rings between $z\in[-4,1]$. In cylindrical coordinates with $z= 1-r^2$, it can be expressed as,

$$\int_S 2\pi r \sqrt{dr^2+dz^2}=2\pi\int_0^{\sqrt5}r\sqrt{1+(z_r')^2}dr =2\pi\int_0^{\sqrt5}r\sqrt{1+4r^2}dr=\frac\pi6(21^{3/2}-1)$$

Note the upper limit of $r$ is $\sqrt5$, not $5$.