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## Homework Statement

Find the surface area of the portion of the sphere [tex]x^2 + y^2 + z^2 = 3c^2[/tex] within the paraboloid [tex]2cz = x^2 + y^2[/tex] using spherical coordinates. (c is a constant)

## Homework Equations

## The Attempt at a Solution

I converted all the x's to [tex]\rho sin\phi cos\theta[/tex], y's to [tex]\rho sin\phi sin\theta[/tex], z's to [tex]\rho cos\theta[/tex]. Thus, the vector form of the sphere is [tex]\vec{r}(\varphi,\theta) = {\rho sin\phi cos\theta}\hat{i} + {\rho sin\phi sin\theta}\hat{j} + {\rho cos\theta}\hat{k}[/tex]. Thus, [tex]{\left\|{{\vec{r}}_{\varphi}}\times {{\vec{r}}_{\theta}}\right\|} = {\sqrt{3c^2}} sin \varphi[/tex]

Now I want to find the bounds of integration for the surface integral. I know theta goes from 0 to [tex]2\pi[/tex], but for [tex]\varphi[/tex], I get this mess while trying to equate the paraboloid to the sphere:

[tex]2c({\rho cos\theta}) = {{\rho}^2 sin^2 \varphi}[/tex]

When [tex]\theta = 0, \varphi = sin^{-1}({\sqrt{2c/{\rho}}})[/tex], and when [tex]\theta = 2\pi, \varphi = sin^{-1}({\sqrt{2c/{\rho}}})[/tex].

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I really hate vector calculus.