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So I was told to find the surface integral $$\iint_{S} yz \mathrm dS$$ for the surface $S$ parametized as $$x=u^2,y=u \sin{v}, z=u \cos{v}$$ for the region $0\leq u\leq1$ and $0\leq v \leq \pi/2$.

So whati did was I computed the magnitude of the cross product, and I got $$\left|\mathbf{r_u} \times \mathbf{r_v}\right|=u\sqrt{1+4u^2}$$ which meant that I had to solve the double integral $$\int_{0}^{1}\int_{0}^{\pi/2}u^2\sin{v}\cos{v} \cdot u\sqrt{1+4u^2}\space \mathrm d u \mathrm d v.$$

If I split the integral using Fubini's Theorem, I'd then evaluate $$\int_{0}^{1}\sin{v}\cos{v} \space\mathrm d v\int_{0}^{\pi/2}u^2 \cdot u\sqrt{1+4u^2} \space \mathrm du.$$

Is this approach correct?

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  • $\begingroup$ Is this a surface integral of a scalar field (as suggested by the function $yz$) or of a vector field (as suggested by the dot product with $\mathrm d \mathbf{S}$? You seem to be mixing both. $\endgroup$ – StackTD Apr 9 '18 at 13:33
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Elaborating on my comment:

So I was told to find the surface integral $$\iint_{S} yz \cdot \mathrm d \mathbf{S}$$

Is this a surface integral of a scalar field (as suggested by the function $yz$) or of a vector field (as suggested by the dot product with the vector $\mathrm d \mathbf{S}$)?

You seem to be mixing both but I'm assuming the first. You would then usually write it as: $$\iint_{S} yz \;\mathrm d S$$ In the general case, for a function $f(x,y,z)$ and a parametrization $\mathbf{r}(u,v)$, you evaluate it as: $$\iint_{S} f(x,y,z) \;\mathrm d S=\iint_{D} f(\mathbf{r}(u,v)) \left|\mathbf{r}_u \times \mathbf{r}_v\right|\,\mathrm d u\,\mathrm d v$$ which is exactly what you did so that looks fine!

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  • $\begingroup$ Thanks for the clarification; I think it's a scale field. I'll edit the question to reflect this $\endgroup$ – user98937 Apr 9 '18 at 14:12
  • $\begingroup$ Also, are the calculations and final integral correct? $\endgroup$ – user98937 Apr 9 '18 at 17:12
  • $\begingroup$ @user98937 Yes, looks good! $\endgroup$ – StackTD Apr 9 '18 at 17:21

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