Surface Area of a Sphere Over a Cone

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So I was reading this thread on the surface integral of a sphere over a cone, but I'm still not sure I understand everything discussed in it. All my functions are the same as the linked thread with different constants. The thread references yet another thread, but the main point of my confusion is: why is the equation inside the integral effectively one. In this example on the linked thread $$\iint _R\sqrt{f_x^2+f_y^2+1}dA$$the value inside the square root is just the Jacobian when youre doing a surface integral. Should we not have to account for the function we are given to integrate over? Furthermore, I'm not sure how to take into account the constraint that it must lie above the cone. How is this done? Any and all help appreciated.

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why is the equation inside the integral effectively one

By this do you mean why is the integrand just the Jacobian $\sqrt{1+f_x^2+f_y^2}$, and not something more? Whenever you are computing arc length, or area, or volume, the function you integrate is just $1$. If there was another function in the integral, you wouldn't be computing the surface area.

Should we not have to account for the function we are given to integrate over?

I'm not 100% sure what you mean by "the function we are given to integrate over". But because of the phrase "over the cone" in the title of the post, I'm guessing you mean the equation defining the cone. This will not enter into the integrand, but will determine the bounds on the integral.

I'm not sure how to take into account the constraint that it must lie above the cone. How is this done?

It depends on what coordinates you're going to use. For instance, if you want to use spherical coordinates, notice that the equation for a cone is always of the form $\phi = \mathrm{const}$. In particular, the cone $z=\sqrt{x^2+y^2}$ is $\phi = \frac{\pi}{4}$ (because it makes a 45 degree angle with the $z$-axis). The points above this cone have smaller $\phi$-values. So the range for $\phi$ is between 0 and $\pi/4$.