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

  • $\begingroup$ 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. $\endgroup$
    – BenCWBrown
    Nov 9 '18 at 2:46
  • $\begingroup$ 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 $\endgroup$
    – mmmmo
    Nov 9 '18 at 4:47
  • $\begingroup$ Okay that makes sense now - I have provided an answer! $\endgroup$
    – BenCWBrown
    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.


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