0
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.