SHORT ANSWER

1 answer:

7 0

(-5,1)(6,-2)
slope = (-2-1) / (6-(-5) = -3/(6 + 5) = -3/11

y = mx + b
slope(m) = -3/11
(-5,1)...x = -5 and y = 1
now we sub and find b, the y int
1 = -3/11(-5) + b
1 = 15/11 + b
1 - 15/11 = b
11/11 - 15/11 = b
-4/11 = b
equation in slope int form : y = -3/11x - 4/11
3/11x + y = - 4/11....multiply by 11
3x + 11y = -4 <== standard form

(4,-2)(4,8).....if u will notice, both the x values are the same. This means the lines are vertical with an undefined slope. Your equation is : x = 4.

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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

ikadub [295]

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

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

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

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