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4vir4ik [10]
3 years ago
6

the length of a rectangle is 4 inches greater than the width. The perimeter of the rectangle is 24 inches what are the dimension

s of the rectangle
Mathematics
1 answer:
deff fn [24]3 years ago
6 0
Perimeter P=2L+2W
P=24
L=w+4
P=2(w+4)+2w
24=2w+8+2w
24=8+4w
16=4w
w=4
L=8
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Because I've gone ahead with trying to parameterize S directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over S straight away, let's close off the hemisphere with the disk D of radius 9 centered at the origin and coincident with the plane y=0. Then by the divergence theorem, since the region S\cup D is closed, we have

\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV

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\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0

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\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0

\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}

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