Finding Surface Area of Sphere Above Cone

[UrbanXrisis]

Surface area integral

sorry, this is not about flux integration... but surface area! sorry about the title!

Find the surface area of the part of the sphere $$x^2+y^2+z^2=36$$ that lies above the cone $$z=\sqrt{x^2+y^2}$$

$$z=\sqrt{36-x^2-y^2}$$


$$A(S)=\int\int_D \sqrt{1+\left( \frac{\partial z}{\partial x} \right) ^2 + \left( \frac{\partial z}{\partial y} \right) ^2 } dA$$

$$\frac{\partial z}{\partial x} =\frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2x \right)$$

$$\frac{\partial z}{\partial y}=\frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2y \right)$$

$$A(S)=\int\int_D \sqrt{1+\left( \frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2x \right) \right) ^2 + \left( \frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2y \right) \right) ^2 } dA$$

is this correct so far? how would I find the ends of integration for a cone?

I've taken the liberty of changing the title of this thread. Since it wasn't even "flux integration" it was bothering me!

 

[Hyperreality]

It would be much easier if you convert to polar coordinates

$$x=r\cos\theta$$

$$y=r\sin\theta$$

and $$z=r$$

Remember to use the Jacobian when you're changing coordinates.

 

[UrbanXrisis]

oh wow, okay, i got it in polar form...

$$\int \int \sqrt{1+r^2 (36-r)^3}$$

$$\int \int \sqrt{1+46656x^2-3888x^4+108x^6-x^8}$$

but what would be the ends of integration for a cone?

 

[0rthodontist]

First find the intersection between the cone and the sphere. This will give you a condition that you can turn into your limits of integration.

 

[UrbanXrisis]

how would I do that?

here's my guess:

the cone has equation:
$$z^2=x^2+y^2$$

the sphere has equation:
$$x^2+y^2+z^2=36$$

putting them together:
$$x^2+y^2=18$$

so my ends would be:

$$\int _0 ^{2\pi} \int_0 ^{\sqrt{18}$$

also, when i change from $$A(S)=\int\int_D \sqrt{1+\left( \frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2x \right) \right) ^2 + \left( \frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2y \right) \right) ^2 } dxdy$$ to the polar form $$\int \int \sqrt{1+r^2 (36-r)^3}dA$$ do I have to add the variable r at the end such that $$\int \int \sqrt{1+r^2 (36-r)^3}rdrd \theta$$?

 

[siddharth]

It should be
$$\int \int \sqrt{1+r^2 (36-r^2)^3} rdrd \theta$$

Apart from that, I think everything else is correct.

 

[vwind]

how the heck do you integrate that?

 

[nizi]

may I suggest this?
$$A(S) = \int\int_S dA = \int_u \int_v \left| { \frac{\partial \mathop r\limits^ \to}{\partial u} \times \frac{\partial \mathop r\limits^ \to}{\partial v} } \right| du dv$$
now considering spherical coordinates, and representing the integral surface with respect to spherical coordinates
$$S : x = 6 \sin \theta \cos \phi , y = 6 \sin \theta \sin \phi , z = 6 \cos \theta$$
$$0 \leq \theta \leq \frac{\pi}{4}, 0 \leq \phi \leq 2 \pi$$
so
$$A(S) = \int_u \int_v \left| { \frac{\partial \mathop r\limits^ \to}{\partial u} \times \frac{\partial \mathop r\limits^ \to}{\partial v} } \right| du dv = \int_\theta \int_\phi \left| { \frac{\partial \mathop r\limits^ \to}{\partial \theta} \times \frac{\partial \mathop r\limits^ \to}{\partial \phi} } \right| d \theta d \phi = \int_\theta \int_\phi 36 \left| { \sin \theta } \right| d \theta d \phi = 36 \int_\theta \left| { \sin \theta } \right| d \theta \int_\phi d \phi =72 \pi \left( { 1 - \frac{1}{\sqrt{2} } \right)$$

 

[vwind]

What is that technique called?
haven't seen that before

 

[Kudaros]

Multiple integrals, usually taught in a third calculus course. Typically comes with a few conversion ideas, like polar coordinates in this example. Each integral and its limits correspond to a variable being integrated.

 

[vwind]

cool, I guess we'll learn that in my next course