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
Best Answer
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.