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Evaluate $$ \iint_s (4x \hat i - 2y^2 \hat j + z^2 \hat k)\cdot \hat n ds $$ over the curved surface of $x^2 + y^2 = 4$ and $z = 0 \text{ to }z = 3$. Using method $$ \iint_s f(x,y(x,z),z)\cdot \frac{\nabla u(x, y)}{|\nabla u(x, y)|} \sqrt{ 1 + \left ( \partial y \over \partial x\right )^2 + \left ( \partial y \over \partial z\right )^2} dxdz$$I got $48 \pi - 128.$
EDIT:: added method for above $$ \int_0^3 \int_{-2}^2 (4x \hat i - 2y^2 \hat j + z^2 \hat k)\cdot(\hat i x + \hat iy) \left ( \sqrt{ 1 + \frac{x^2}{4 - x^2 } }\right ) dx dz$$ $$ \implies 12 \int_{-2}^{2} \frac{2x^2}{\sqrt{4 - x^2}} - (4 - x^2)dx = 48 \pi - 128 $$ EDIT:: I got wrong answer below because of formula. I got right answer from this! The correct formula should have been $$ \iint_s F(\theta, z) \cdot (u_\theta \times u_z) d\theta dz$$ Parametrizing $x = 2 \cos \theta, y = 2\sin\theta , z=z $ $$ \int_0^3 \int_0^{2\pi} F(\theta, z)\cdot \frac{u_{\theta}\times u_z}{|u_{\theta}\times u_z|}d\theta dz$$ I got $24 \pi$. The answer sheet says $48\pi$.Please help!! Thank you!!

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Can you show a little detail about the computations? We can't tell where you go wrong if we don't see where you went. =) – Patrick Da Silva Jun 22 '12 at 9:27
@PatrickDaSilva does it help you to help me?? – Santosh Linkha Jun 22 '12 at 9:42
I was asleep, so for a while, it didn't, but it certainly helped you to help yourself! – Patrick Da Silva Jun 22 '12 at 15:39
up vote 4 down vote accepted

The quantity $\Phi$ (for flux) you want to compute is given by the formula $$\Phi=\int_R {\bf F}\bigl({\bf u}(\theta, z)\bigr) \cdot ({\bf u}_\theta \times {\bf u}_z)\ {\rm d}(\theta ,z)\ ,\qquad R:=[0,2\pi]\times[0,3]\ ,$$ which you quote correctly. But in the next line for no reason you divide by $ \bigl|{\bf u}_\theta \times {\bf u}_z\bigr|=2$; therefore your end result is off by a factor of $2$.

As an aside: The orientation of ${\bf n}$ was not defined; so maybe the intended value is $-48\pi$.

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I got that myself ... could you please view at my first method ... both method are right? – Santosh Linkha Jun 22 '12 at 15:15

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