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Evaluate the flux integral $\displaystyle \int \int_S {\bf F \cdot n} \ dS$ Where ${\bf F}(x,y,z) = zy {\bf j}$ is the portion of the paraboloid $z = 1 - x^2 -y^2$ above the xy-plane such that also $x \geq 0$.

We derived the formula: $\displaystyle \int \int_S {\bf F \cdot n} \ dS = \int \int_R (-F_1 f_x - F_2 f_y + F_3) \ dx \ dy$ Applying this I get $\displaystyle \int \int_S {\bf F \cdot n} \ dS = \int \int_R 2zy^2 \ dx \ dy = \int \int_R 2(1-r^2)(r^2\sin^2 \theta)r \ dr \ d\theta$,

however, I'm having troubles finding the limits of r and theta, in examples I see online, they just take $ 0 \leq r \leq 1$ and $0 \leq \theta \leq 2\pi$, but I don't understand why, also - why is the $x \geq 0 $ relevant here? I don't see how I could project it and get limits for the x,y plane in this question

any help please

Warz
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1 Answers1

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First you have to consider only the projection of your surface on $xy$ plane i.e. $$z=0\implies 0=1-x^2-y^2,x\ge 0$$ which is a semicircle and so the limits are$$r:0\to 1,\theta:-\pi/2\to\pi/2$$

Semsem
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  • is there anyway you could derive the limits without knowing how the projection looks like? – Warz Mar 02 '14 at 15:03
  • No, this is the region you have to integrate over. – Semsem Mar 02 '14 at 15:09
  • thank you - one last question, if we instead (for whatever reason), wanted to project the graph $f(x,y)$ in say, the x-z plane - how would the formula change for the flux integral? – Warz Mar 02 '14 at 15:17
  • you have to define the surface as $y=g(x,z)$ and to get $(F\cdot n)dS$ as you did before. in this case the projection will be the region bounded by the parabola $z=1-x^2, x\ge0$ – Semsem Mar 02 '14 at 15:23
  • would the formula then involve $g_x $ and $g_z$ rather than $f_x$ and $f_y$? (partial derivative) – Warz Mar 02 '14 at 15:25
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    $\displaystyle \int \int_S {\bf F \cdot n} \ dS = \int \int_{R_{xz}} (-F_1 g_x - F_2 g_z + F_3) \ dx \ dz$ – Semsem Mar 02 '14 at 15:31