Surface area of a sphere

• Jul 31st 2009, 07:24 AM
Surface area of a sphere
Find the area of the portion of the sphere of radius 2 (centered at the origin) that is in the cone $\displaystyle z > \sqrt {x^2+y^2}$

What I have so far is that:

$\displaystyle A(S)= \int \int \int \sqrt {1+ \frac {x^2}{x^2+y^2}+\frac {y^2}{x^2+y^2} } dA$

=$\displaystyle \int \int \int \sqrt {2} dA$
=$\displaystyle \sqrt {2} \int ^2 _0 \int ^{2 \pi } _0 \int ^ { \pi } _0 p^2 sin \phi d \phi d \theta dp$
Then I have $\displaystyle \frac {16 sqrt {2} \pi }{3}$ But is wrong, what is wrong?

Thank you.
• Jul 31st 2009, 08:03 AM
arbolis
Quote:

$\displaystyle A(S)= \int \int \int \sqrt {1+ \frac {x^2}{x^2+y^2}+\frac {y^2}{x^2+y^2} } dA$
It is not true I believe. It should be a double integral since you have $\displaystyle dA$.
Take it from here, I think all should go well.
• Jul 31st 2009, 08:46 AM
Haytham
maybe you could try to find the cone generating angle $\displaystyle \theta_0$ and then calculate the integral with $\displaystyle \theta=0..\theta_0$.

no need include anything else in the integral besides $\displaystyle dA$
• Jul 31st 2009, 10:10 AM
Jester
Quote:

Find the area of the portion of the sphere of radius 2 (centered at the origin) that is in the cone $\displaystyle z > \sqrt {x^2+y^2}$

What I have so far is that:

$\displaystyle A(S)= \int \int \int \sqrt {1+ \frac {x^2}{x^2+y^2}+\frac {y^2}{x^2+y^2} } dA$

=$\displaystyle \int \int \int \sqrt {2} dA$
=$\displaystyle \sqrt {2} \int ^2 _0 \int ^{2 \pi } _0 \int ^ { \pi } _0 p^2 sin \phi d \phi d \theta dp$
Then I have $\displaystyle \frac {16 sqrt {2} \pi }{3}$ But is wrong, what is wrong?

Thank you.

Quote:

Originally Posted by arbolis
It is not true I believe. It should be a double integral since you have $\displaystyle dA$.
Take it from here, I think all should go well.

In addition, I think you should be using

$\displaystyle z = \sqrt{4-x^2-y^2}$ in your surface area formula

$\displaystyle \iint_R \sqrt{1 + z_x^2 + z_y^2}\,dA.$
• Jul 31st 2009, 11:22 AM
Opalg
As Danny says, you should write the equation of the sphere as $\displaystyle z = \sqrt{4-x^2-y^2}$. To find the region over which to integrate, you need to know where the sphere meets the cone, namely where $\displaystyle \sqrt{x^2+y^2} = \sqrt{4-x^2-y^2}$ or in other words $\displaystyle x^2+y^2=2$. So the integral will be over the region $\displaystyle x^2+y^2\leqslant2$, and the surface area is given by the integral $\displaystyle \iint_{x^2+y^2\leqslant2}\sqrt{1+z_x^2+z_y^2}dxdy = \iint_{x^2+y^2\leqslant2}\frac2{\sqrt{4-x^2-y^2}}dxdy$. This cries out for conversion to polar coordinates, and it then becomes $\displaystyle \int_0^{2\pi}\!\!\int_0^{\sqrt2}\!\!\frac2{\sqrt{4-r^2}}\,rdrd\theta$. From there on, it's easy.
• Aug 2nd 2009, 10:12 AM
alan collins
Quote:

Originally Posted by Opalg
As Danny says, you should write the equation of the sphere as $\displaystyle z = \sqrt{4-x^2-y^2}$. To find the region over which to integrate, you need to know where the sphere meets the cone, namely where $\displaystyle \sqrt{x^2+y^2} = \sqrt{4-x^2-y^2}$ or in other words $\displaystyle x^2+y^2=2$. So the integral will be over the region $\displaystyle x^2+y^2\leqslant2$, and the surface area is given by the integral $\displaystyle \iint_{x^2+y^2\leqslant2}\sqrt{1+z_x^2+z_y^2}dxdy = \iint_{x^2+y^2\leqslant2}\frac2{\sqrt{4-x^2-y^2}}dxdy$. This cries out for conversion to polar coordinates, and it then becomes $\displaystyle \int_0^{2\pi}\!\!\int_0^{\sqrt2}\!\!\frac2{\sqrt{4-r^2}}\,rdrd\theta$. From there on, it's easy.

What about if the sphere were radius 5? I seem to be goofing it up just a little somewhere when converting to polar/spherical coordinates. Also, it seems as though the Jacobian should be (p^2)sin(phi) instead of just r because this should be changed to polar and not cylindrical coordinates... right? Any feedback, insight or explanation of this integral problem would be fantastic as I am seriously stuck and puzzled by this problem. Thanks.
• Aug 2nd 2009, 03:02 PM
Haytham
I think it would be easier to do the all thing in spherical coordinates.

Notice that if the height of the cone is z (a generic point z), and the radius of the base is $\displaystyle \sqrt{x^2+y^2}$, so the generating angle for the cone, $\displaystyle \theta_0$, is given by $\displaystyle tan \theta_0=\frac{\sqrt{x^2+y^2}}{z}$

as $\displaystyle z=\sqrt{x^2+y^2}$, $\displaystyle tan \theta_0=1$ and $\displaystyle \theta_0=\frac{\pi}{4}$

this makes the integral for the area

$\displaystyle \int_0^{2\pi}\int_0^{\frac{\pi}{4}}R^2sin\theta d\theta d\phi$

it seems much more simpler to me, and no need to change the coordinates in between. the R being 2 or 5 makes no difference you just replace it in the integral as a constant.