0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ @Nosrati no sir, it is $4r^4(cos^2 \theta + sin^2 \theta)+r^2$ $\endgroup$ – Magneto Aug 19 '18 at 20:20
  • $\begingroup$ no problem sir, ..... $\endgroup$ – Magneto Aug 19 '18 at 20:24
  • $\begingroup$ Your first integral $\iint_S (y^2z dxdy+xzdydz+x^2ydxdz)$ is the victim of some private notation used by your textbook or teacher. It does not make sense to me. $\endgroup$ – Christian Blatter Aug 20 '18 at 10:54
  • $\begingroup$ @ChristianBlatter i understood sir, below post explains everthing. This is textbook notation. This is exercise problem $\endgroup$ – Magneto Aug 20 '18 at 13:24
1
$\begingroup$

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 $$\int _0^{\frac{\pi }{2}}\int _0^1\left(z \cos ^2(t)+\sin ^2(t) \cos ^2(t)\right)dzdt=\dfrac{3\pi}{16}$$

$\endgroup$
2
  • $\begingroup$ Sir now it is perfect. $-(\pi/16)+(3 \pi /16) = (\pi/8)$. I did by gauss divergence theorem. It is straight forward and it is also giving $\pi/8$. Thanks a lot. $\endgroup$ – Magneto Aug 20 '18 at 7:48
  • $\begingroup$ Good luck $;)$. $\endgroup$ – Nosrati Aug 20 '18 at 8:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.