# 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

Although the sine substitution probably is the easiest method, here's another one:

If the integrand is a fraction with a square root in the denominator, see if you can write the integrand as the derivative of the square root times another function to pave the way for an integration by parts:

$I = \int \frac{r^3}{\sqrt{4-r^2}}\, \mathrm{d}r = \int -r^2 \frac{-r}{\sqrt{4-r^2}}\, \mathrm{d}r$

$\: = -r^2 \sqrt{4-r^2} + 2 \int r \sqrt{4-r^2}\, \mathrm{d}r$

$\: = -r^2 \sqrt{4-r^2} - \frac{2}{3} (4-x^2)^{\frac{3}{2}}$

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