Find the value of Surface Integral $ \displaystyle \int \int yzdxdy + xzdydz + xydxdz$ Where the surface is the outer side of the surface formed by the cylinder $x^2 + y^2 =4 $ and the plane $x = 0 , y= 0 ,z = 0$ and $z= 2$
Now I would like to solve this surface integral with the help of Gauss Divergence theorem, but firstly I want the surface integral in the form $\int F.n ds$
so that I can calculate $\nabla. F$
Can anyone tell me how should I solve this integral ?
Thank you .
This is a surface integral of the second kind. A surface integral of the form
$$\iint_R Pdydz + Qdxdz + Rdxdy$$ $$= \iint_S (Pcos\alpha + Qcos\beta + Rcos\gamma)dS$$
Where
$$cos\alpha = \frac{A}{\sqrt{A^2 + B^2 + C^2}}$$ $$cos\beta = \frac{B}{\sqrt{A^2 + B^2 + C^2}}$$ $$cos\gamma = \frac{C}{\sqrt{A^2 + B^2 + C^2}}$$
Where $$A = \frac{\partial{(y, z)}}{\partial{(u, v)}}$$ $$B = \frac{\partial{(z, x)}}{\partial{(u, v)}}$$ $$C = \frac{\partial{(x, y)}}{\partial{(u, v)}}$$
Based on these formulae we use the following parameter for the cylinder $$x = 2cos\theta $$ $$y = 2sin\theta $$ $$z = z$$
Using the above formulae we have $$\sqrt{A^2 + B^2 + C^2} = 2$$ Where $$A = 2cos\theta$$ $$B = 2sin\theta $$ $$C = 0$$ Substitute x, y, z into the integral we have $$I = \iint_S (2zsin\theta cos\theta + 2zcos\theta sin \theta)dS$$
Using the change of variables from x, y, z to $\theta$ and z, J factor is 2.
$$= \int_0^{2}\int_0^{\frac{\pi}{2}} 4z sin \theta cos \theta d\theta dz= 4$$