Compute the integral $\iint_S (y^2z dxdy+xzdydz+x^2ydxdz)$ where S is the outer side of the surfaces situated in the first octant and formed by the paraboloid of revolution $z=x^2+y^2, $ cylinder $x^2+y^2=1$ and the coordinate planes
My attempt:-
$F=<xz,x^2y,y^2z>, \vec n = \frac{2x \vec i+2y \vec j- \vec k}{\sqrt{4(x^2+y^2)+1}}\\ \vec F \cdot \vec n = \frac {2 x^2z+2x^2y^2-y^2z}{\sqrt{4(x^2+y^2)+1}} \\ \iint_S \vec F \cdot \vec n dS \\ = \iint \frac {2 x^2z+2x^2y^2-y^2z}{\sqrt{4(x^2+y^2)+1}} dS$
Taking $\vec R = (r \cos \theta, r \sin \theta, r^2) \\\vec R_r = (\cos \theta, \sin \theta, 2r) \\\vec R_{\theta} = (-r \sin \theta,r \cos \theta, 0) \\ \vec R_r \times \vec R_{ \theta} = \vec i(-2r^2 \cos \theta ) – \vec j (2r^2 \sin \theta) + \vec k (r) \\ |\vec R_r \times \vec R_{ \theta} | = r \sqrt{4r^2+1}$
$\iint_S \vec F \cdot \vec n dS = \int_{\theta=0}^{\pi/2} \int_{r=0}^1 (2r^5 \cos^2 \theta+2r^5 \sin^2 \theta \cos^2 \theta – r^5 \sin^2 \theta) dr d \theta = \fbox {$\frac{\pi}{16}$}$
But given answer is $\fbox {$\frac{\pi}{8}$}$
I used gauss divergence theorem, i am getting $\fbox {$\frac{\pi}{8}$}$
Pls tell me where i went wrong.
Best Answer
Hint: I think you droped other surfaces. One may prove for $x=0$, $y=0$ and $z=0$ the integrals are zero. The direction of the surface paraboloid $z=x^2+y^2$ is outer side, that is $$\vec n = \color{red}{-}\frac{2x \vec i+2y \vec j- \vec k}{\sqrt{4(x^2+y^2)+1}}$$ also consider cylinder $x^2+y^2=1$ with parametrization $$\vec R = (\cos \theta, \sin \theta, z)$$ gives $$\int _0^{\frac{\pi }{2}}\int _0^1\left(z \cos ^2(t)+\sin ^2(t) \cos ^2(t)\right)dzdt=\dfrac{3\pi}{16}$$