## AakashSudhakar one year ago [Calculus 3] Can someone help me with the following surface integral? Thanks!

1. AakashSudhakar

Evaluate the surface integral $\int\limits_{ }^{ }\int\limits_{S}^{}(x^2+y^2+5z^2)dS$ where S is the part of the cone $z=6-\sqrt{x^2+y^2}$ above the plane $z=0$

2. IrishBoy123

use cylindrical so $$z = 6 - r$$ or: $$z +r-6=0$$ and area element $$\large dS = \frac{r dr d \theta}{cos \alpha}$$ where $$\large cos \alpha = \frac{\vec n . \hat z}{|\vec n | \ | \hat z|}$$ [ie angle between normal to cone and the z axis] normal $$\vec n = \nabla [z+r-6] = <1,0,1>$$, $$cos \alpha = \frac{1}{\sqrt{2}}$$ [meets reality test, cone at $$45^o$$ to z axis] the integrand: $$x^2+y^2+5z^2 = r^2 + 5(6-r)^2 = r^2 + 5(36 - 12r + r^2)) \\ = 180 - 60r + 6r^2$$ $$\large = \sqrt{2} \int_{\theta=0}^{2\pi} \int_{r=0}^{6} (180 - 60r + 6r^2)r \ dr \ d \theta$$

3. AakashSudhakar

Thank you, I could not for the life of me figure this out, particularly regarding solving for the normal vector and using the dot product formula for cosine. That didn't even occur to me. Thanks for all the help!