# calculate $\iint z dS$ where S is the upper hemisphere of radius a.

I came across the following problem in my textbook and my answer differs from the one given and I just wanted to check my work to see where I went wrong

calculate $$\iint z dS$$ where S is the upper hemisphere of radius a.

So first I set $$z = \sqrt(a-x^2-y^2)$$

so $$dz/dx$$ = $$-x/\sqrt(a-x^2-y^2)$$ and $$dz/dx$$ = $$-y/\sqrt(a-x^2-y^2)$$

Thus this integral becomes $$\iint adS$$

Changing to polar coordinates we get $$\iint a^3sin\phi d\phi d\theta$$ for $$\theta$$ between $$0$$ and $$2\pi$$ and $$\phi$$ between 0 and $$\pi/2$$. Computing this integral we get that it gives $$2\pi a^3$$.

However, the answer provided is $$\pi a^3$$. Is there an error in my calculations? Or is the textbook provided answer incorrect? Thanks

• I am not quite sure how you arrive at $\iint adS$ from your working, but I notice that you're missing $a^{2}$ in $z = \sqrt{a^{2} - x^{2} - y^{2} }$. It would be simpler to work in spherical polar coordinates $(x,y,z) = (a\sin\theta\cos\phi, a\sin\theta\sin\phi, a\cos\theta)$ from the beginning. I did the question that way and arrived at $\iint a^{3}\sin\phi\cos\phi d\phi d\theta$, which resulted in the provided answer. Nov 9 '18 at 2:46
• I used the fact that we are integrating over a graph and thus the integral can be simplified to f(x,y,z)$\sqrt(1+(dz/dx)^2+(dz/dy)^2)$. Then this simplifies to z*a/z if I am not mistaken Nov 9 '18 at 4:47
• Okay that makes sense now - I have provided an answer! Nov 9 '18 at 5:43

Let $$S$$ be the surface of the hemisphere of radius $$a$$, and let $$D$$ be the disk underneath it in the $$xy$$-plane. Then we can consider the hemisphere as the graph of a function $$f(x, y, z(x,y)) = z(x,y)$$, where $$z(x,y) = \sqrt{a^{2} - x^{2} - y^{2}}$$. The surface integral of $$S$$ is: $$\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{f\left( {x,y,z\left( {x,y} \right)} \right)\sqrt {{{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2} + 1} \,dA}},$$ where $$dA$$ is the area element of $$D$$. Then like you calculated, $$\frac{\partial z}{\partial x} = \frac{-x}{\sqrt{a^{2}-x^2-y^2}} = -\frac{x}{z},\qquad \frac{\partial z}{\partial y} = \frac{-y}{\sqrt{a^{2}-x^2-y^2}} = -\frac{y}{z},$$ so $$\sqrt {{{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2} + 1} = \frac{1}{z}\bigg(x^{2} + y^{2} + z^{2}\bigg)^{\tfrac{1}{2}} = \frac{a}{z}.$$ Substituting this in yields $$\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{z\cdot\frac{a}{z} \,dA}} = a\iint\limits_{D}{dA}.$$ In spherical polar coordinates, $$(x,y) = (a\sin\theta\cos\phi, a\sin\theta\sin\phi)$$, for $$0 \leq \theta < \tfrac{\pi}{2}$$, $$0 \leq \phi < 2\pi$$. So as $$dA = dx\, dy$$, then $$dA = dx\, dy = \det\frac{\partial (x, y)}{\partial(\theta,\phi)} d\theta\, d\phi = a^{2}\sin\theta\cos\theta d\theta\, d\phi = \frac{a^{2}}{2} \sin(2\theta) d\theta\, d\phi,$$ and consequently $$a\iint\limits_{D}{dA} = \frac{a^{3}}{2}\int_{0}^{\frac{\pi}{2}} \sin(2\theta) d\theta\, \int_{0}^{2\pi} d\phi = \frac{a^{3}}{2} \cdot \bigg[\frac{-1}{2}(-1 -1 )\bigg] \cdot 2\pi = a^{3}\pi,$$ as the expected answer should be. Comparing this to your attempt, it looks like the mistake occurs when changing from the Cartesian area element to the spherical polar are element.