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morpeh [17]
3 years ago
11

Calculate the flux of the vector field F⃗ (x,y,z)=(exy+9z+4)i⃗ +(exy+4z+9)j⃗ +(9z+exy)k⃗ through the square of side length 3 wit

h one vertex at the origin, one edge along the positive y-axis, one edge in the xz-plane with x≥0 and z≥0, oriented downward with normal n⃗ =i⃗ −k
Mathematics
1 answer:
ikadub [295]3 years ago
3 0

The square (call it S) has one vertex at the origin (0, 0, 0) and one edge on the y-axis, which tells us another vertex is (0, 3, 0). The normal vector to the plane is \vec n=\vec\imath-\vec k, which is enough information to figure out the equation of the plane containing S:

(x\,\vec\imath+y\,\vec\jmath+z\,\vec k)\cdot(\vec\imath-\vec k)=0\implies x-z=0\implies z=x

We can parameterize this surface by

\vec s(x,y)=x\,\vec\imath+y\,\vec\jmath+x\,\vec k

for 0\le x\le\frac3{\sqrt2} and 0\le y\le3. Then the flux of \vec F, assumed to be

\vec F(x,y,z)=(e^{xy}+9z+4)\,\vec\imath+(e^{xy}+4z+9)\,\vec\jmath+(9ze^{xy})\,\vec k,

is

\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\iint_S\vec F(\vec s(x,y))\cdot\vec n\,\mathrm dx\,\mathrm dy

=\displaystyle\int_0^3\int_0^{3/\sqrt2}\left((4+e^{xy}+9x)\,\vec\imath+(9+e^{xy}+4x)\,\vec\jmath+(e^{xy}+9x)\,\vec k\right)\cdot(\vec\imath-\vec k)\,\mathrm dx\,\mathrm dy

=\displaystyle\int_0^3\int_0^{3/\sqrt2}4\,\mathrm dx\,\mathrm dy=\boxed{18\sqrt2}

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