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3 years ago

Please help this is important

Mathematics

1 answer:

deff fn [24]3 years ago

5 0

Answer:

below

Step-by-step explanation:

You've been asked to use specific x-values. Simply replace the x in the equation for each of your x-values then graph.

first equation!!

(x,y)

y=-2x+5

x-value: -2

y=-2(-2)+5

y=4+5

y=9

(-2, 9)

x-value: -1

y=-2(-1)+5

y=2+5

y=7

(-1, 7)

x-value: 0

y=-2(0)+5

y=0+5

(0, 5)

x-value: 1

y=-2(1)+5

y=-2+5

y=3

(1, 3)

x-value: 2

y=-2(2)+5

y=-4+5

y=1

(2, 1)

second equation :)

(x,y)

y=x^2 +2

x-value: -2

y=-2^2 +2

y=4+2

y=6

(-2, 6)

x-value: -1

y=-1^2 +2

y=1+2

y=3

(-1, 3)

x-value: 0

y=0^2 +2

y=0+2

(0, 2)

x-value: 1

y=1^2 +2

y=1+2

y=3

(1, 3)

x-value: 2

y=2^2 +2

y=4+2

y=6

(2, 6)

third equation

(x,y)

y=√2 x+1

x-value: 1

y=√2 1+1

y=1+1

y=2

(1, 2)

x-value: 4

y=√2 4+1

y=2+1

y=3

(4, 3)

x-value: 9

y=√2 9+1

y=3+1

y=4

(9, 4)

x-value: 16

y=√2 16+1

y=4+1

y=5

(16, 5)

x-value: 25

y=√2 25+1

y=5+1

y=6

(25, 6)

fourth (last) equation :-)

(x,y)

y=4/x

x-value: -4

y=4/-4

y=-1

(-4, -1)

x-value: -2

y=4/-2

y=-2

(-2, -2)

x-value: -1

y=4/-1

y=-4

(-1, -4)

x-value: 0

y=4/0

y=0

(0, 0)

x-value: 1

y=4/1

y=4

(1, 4)

x-value: 2

y=4/2

y=2

(2, 2)

x-value: 4

y=4/4

y=1

(4, 1)

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Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of

kifflom [539]

Looks like we have

$\vec F(x,y,z)=z^2x\,\vec\imath+\left(\dfrac{y^3}3+\sin z\right)\,\vec\jmath+(x^2z+y^2)\,\vec k$

which has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(z^2x)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial z}=z^2+y^2+x^2$

By the divergence theorem, the integral of $\vec F$ across $S$ is equal to the integral of $\nabla\cdot\vec F$ over $R$ , where $R$ is the region enclosed by $S$ . Of course, $S$ is not a closed surface, but we can make it so by closing off the hemisphere $S$ by attaching it to the disk $x^2+y^2\le1$ (call it $D$ ) so that $R$ has boundary $S\cup D$ .

Then by the divergence theorem,

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV$

Compute the integral in spherical coordinates, setting

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

so that the integral is

$\displaystyle\iiint_R(x^2+y^2+z^2)\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^1\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{2\pi}5$

The integral of $\vec F$ across $S\cup D$ is equal to the integral of $\vec F$ across $S$ plus the integral across $D$ (without outward orientation, so that

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\iint_D\vec F\cdot\mathrm d\vec S$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le1$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-u\,\vec k$

Then we have

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1\left(\frac{u^3}3\sin^3v\,\vec\jmath+u^2\sin^2v\,\vec k\right)\times(-u\,\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^1u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac\pi4$

Finally,

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\left(-\frac\pi4\right)=\boxed{\frac{13\pi}{20}}$

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3 years ago

Read 2 more answers

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3 years ago

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