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

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

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

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

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

3 0

3 years ago

5x+2y=10 in function notation

Zanzabum

Normally it is x ininput and f(x)=y
solve for y and replace y with f(x)

minus 5x from both sides
2y=10-5x
divide both sides by 2
y=5-2.5x
f(x)=5-2.5x

4 0

3 years ago