Math Help - Surface area of a sphere

1. 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 $z > \sqrt {x^2+y^2}$

What I have so far is that:

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

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

Thank you.

2. $A(S)= \int \int \int \sqrt {1+ \frac {x^2}{x^2+y^2}+\frac {y^2}{x^2+y^2} } dA$
It is not true I believe. It should be a double integral since you have $dA$.
Take it from here, I think all should go well.

3. maybe you could try to find the cone generating angle $\theta_0$ and then calculate the integral with $\theta=0..\theta_0$.

no need include anything else in the integral besides $dA$

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

What I have so far is that:

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

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

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

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

$
\iint_R \sqrt{1 + z_x^2 + z_y^2}\,dA.
$

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

6. What about a shpere with radius 5 in the same question?

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

7. 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 $\sqrt{x^2+y^2}$, so the generating angle for the cone, $\theta_0$, is given by $tan \theta_0=\frac{\sqrt{x^2+y^2}}{z}$

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

this makes the integral for the area

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

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.