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Let $S$ denote the hemisphere $x^2+y^2+z^2=1$ , $z\ge0$, and let $F(x,y,z)=xi+yj$. Let $n$ be the unit outward normal of $S$. Compute the value of the surface integral $\iint_S{F.n} dS$ , using the explicit representation $z=\sqrt{1-x^2-y^2}$

My attempt : I know the formula $\iint_S{F.n} dS= \iint_{T} F.n |\frac{\partial{r}}{\partial u} \times \frac{\partial{r}}{\partial v}|dudv$

But i don't know how to apply the formula in the given question

jasmine
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1 Answers1

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Using divergence theorem, one can easily see that the result is volume of the unit sphere which is $\frac{4\pi}{3}$. But as the question specifically seeks to use representation $z=\sqrt{1-x^2-y^2}$ and do surface integral,

use the fact that radially outward normal vector for a sphere is $(x,y,z)$ which can be rewritten as $(\frac{x}{\sqrt{1-x^2-y^2}}, \frac{y}{\sqrt{1-x^2-y^2}}, 1)$ using the representation $z = \sqrt{1-x^2-y^2}$.

Or you can take partial derivative as below,

$-\displaystyle \frac{\partial z}{\partial x} = \frac{x}{\sqrt{1-x^2-y^2}}$

$-\displaystyle \frac{\partial z}{\partial y} = \frac{y}{\sqrt{1-x^2-y^2}}$

So the normal vector, $\vec n = (\frac{x}{\sqrt{1-x^2-y^2}}, \frac{y}{\sqrt{1-x^2-y^2}}, 1)$

$\vec F = (x,y,0)$

$\vec F \cdot \vec n = \frac{x^2 + y^2}{\sqrt{1-x^2-y^2}}$

Surface integral should be

$\displaystyle \int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \frac{x^2 + y^2}{\sqrt{1-x^2-y^2}} \ dx \ dy$

Math Lover
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    Let me know if any questions. Basically when representing surface as $z = f(x,y)$, we have to make $z$ component $1$ in the normal vector. Also $z = \sqrt{1-x^2-y^2}$ represents a hemisphere with $z \geq 0$. $z = - \sqrt{1-x^2-y^2}$ will represent the hemisphere with $z \leq 0$. – Math Lover Feb 13 '21 at 13:29
  • thanks u @Math Lover but im not able to proceed further can u elaborate more ...this is my attempt :$\displaystyle \int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \frac{x^2 + y^2}{\sqrt{1-x^2-y^2}} \ dx \ dy$

    I put $y= r \sin \theta , x=r \cos\theta$..

    $\displaystyle \int_{\pi}^{2\pi} \int_{-1}^{1} \frac{r^3}{\sqrt{1-r^2}} \ \cos^2\theta dr \ d\theta $

     – jasmine Feb 13 '21 at 14:07
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    If you are substituting $x = r \cos \theta, y = r \sin \theta, 0 \leq r \leq 1$, $0 \leq \theta \leq {2\pi}$, when you convert from cartesian $dx \ dy$ to polar $dr \ d\theta$, the Jacobian is simply $r$. $\cos^2\theta$ should not be there. – Math Lover Feb 13 '21 at 14:17
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    ohh ! got its now @Math lover – jasmine Feb 13 '21 at 14:20