# Really Confused on a surface area integral can't seem to finish the integral off.

Basically the question asks to compute $\int \int_{S} ( x^{2}+y^{2}) dA$ where S is the portion of the sphere $x^{2} + y^{2}+ z^{2}= 4$ and $z \in [1,2]$ we start with a chnage of variables

$x=x$

$y=y$

$z= 2 \cdot(4-(x^{2} + y^{2}))^{1/2}$

$Det(u,v)= \begin{bmatrix} i & j& k \\ 1 & 0 & \frac {-x}{(4-(x^{2} + y^{2}))^{1/2}} \\ 0 & 1 & \frac {-y}{(4-(x^{2} + y^{2}))^{1/2}} \\ \end{bmatrix}=(\frac {x}{(4-(x^{2} + y^{2}))^{1/2}})i + (\frac {y}{(4-(x^{2} + y^{2}))^{1/2}})j + k$

$dA=(\frac {x^{2}+y^{2}}{(4-(x^{2} + y^{2}))} +1)^{1/2}$

$\int \int_{S} (\frac {(x^{2}+y^{2})^{3}}{(4-(x^{2} + y^{2}))}+(x^{2}+y^{2})^{2})^{1/2}$

Projecting when z=1 and z=2 we have $x^{2} + y^{2}= 4-1$ $\to r= 0,(3)^{1/2}$ going to polar we have:

$(\frac {r^{6}}{(4-r^{2})}+r^{4})^{1/2}rdrd\theta=(\frac {4r^{4}}{(4-r^{2})})^{1/2}rdrd\theta$

my problem is $2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} (\frac {4r^{4}}{(4-r^{2})})^{1/2}rdrd\theta$ there is no nice way i can think of to integrate this. it can also be written as:

$2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} \frac {2r^{2}}{((4-r^{2}))^{1/2}}rdrd\theta$

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Why do you have $z = 2 \sqrt{4-x^2-y^2}$? That doesn't seem to satisfy the equation of the sphere. –  Ron Gordon Apr 9 '13 at 21:45
the 2 is jsut to represent 2* the surface area of $(4-x^{2}-y^{2})^{1/2}$ –  Faust7 Apr 9 '13 at 21:56
That doesn't seem right. The parameters must satisfy the equation of the surface; multiply one of them by $2$ throws that off. –  Ron Gordon Apr 9 '13 at 21:58
im not multiplying the value by 2 see the random 2 outside my integral? that where it went because i have twice as much surface area as i have strictly positive surface area by symmetry no? or am i supposed to throw it out since $z>1$ ? –  Faust7 Apr 9 '13 at 22:02
No, I don't see the symmetry you speak of in $z$. –  Ron Gordon Apr 9 '13 at 22:37

Sometimes cylindrical coordinates is at least as easy when there is axial symmetry. The integral to be evaluated becomes

$$2 \pi \int_1^2 dz \, r(z)^3 \sqrt{1+\left ( \frac{d r}{dz} \right)^2}$$

where $r(z) = \sqrt{4 - z^2}$ and $r$ is the distance from the axis to the sphere. This reduces to, upon evaluation of the terms in the integrand

$$4 \pi \int_1^2 dz \: (4-z^2) = \frac{20 \pi}{3}$$

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lol wut? just in case someone didn't follow. $\frac{dr}{dz}= \frac {-z}{4-z^{2}}$ $\to$ $1+\frac {z^{2}}{(4-z^{2})} = \frac {2}{(4-z^{2})^{1/2}} \to \frac {2}{r(z)}$ nice catch m8 –  Faust7 Apr 9 '13 at 21:59

It seems that spherical coordinates are more appropriate for calculating $\iint\limits_{S} ( x^{2}+y^{2}) dA$.
In the last integral $$2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} \frac {2r^{2}}{((4-r^{2}))^{1/2}}rdrd\theta= {4\pi}\int^{(3)^{1/2}}_{0} \frac {2r^{2}}{((4-r^{2}))^{1/2}}r\,dr$$ you can make the substitution $r=2\sin{t}.$

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Thats pretty nasty but ty that work's not pleasantly but the internet has an answer to that integrande. –  Faust7 Apr 9 '13 at 21:40
You might want to verify that the OP's integral is even correct. Evaluating it, I got $64 \pi [(\pi/6) - (\sqrt{3}/8)]$, which makes no sense to me. –  Ron Gordon Apr 9 '13 at 21:44
$I = \int \frac{r^3}{\sqrt{4-r^2}}\, \mathrm{d}r = \int -r^2 \frac{-r}{\sqrt{4-r^2}}\, \mathrm{d}r$
$\: = -r^2 \sqrt{4-r^2} + 2 \int r \sqrt{4-r^2}\, \mathrm{d}r$
$\: = -r^2 \sqrt{4-r^2} - \frac{2}{3} (4-x^2)^{\frac{3}{2}}$