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meriva
3 years ago
13

Use Green's Theorem to calculate the circulation of F =2xyi around the rectangle 0≤x≤8, 0≤y≤3, oriented counterclockwise.

Mathematics
1 answer:
Tamiku [17]3 years ago
6 0

Green's theorem says the circulation of \vec F along the rectangle's border C is equal to the integral of the curl of \vec F over the rectangle's interior D.

Given \vec F(x,y)=2xy\,\vec\imath, its curl is the determinant

\det\begin{bmatrix}\frac\partial{\partial x}&\frac\partial{\partial y}\\2xy&0\end{bmatrix}=\dfrac{\partial(0)}{\partial x}-\dfrac{\partial(2xy)}{\partial y}=-2x

So we have

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_D-2x\,\mathrm dx\,\mathrm dy=-2\int_0^3\int_0^8x\,\mathrm dx\,\mathrm dy=\boxed{-192}

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Find the exact value of sin ( 7 π 4 ) using reference angles
Firdavs [7]

Answer:

-1/root2

Step-by-step explanation:

7pi/4 radians is the same as -pi/4 radians

sin(-pi/4) is -sin(pi/4) = -1/root2

3 0
3 years ago
Two classes have a total of 60 students. The students need to make teams of 8t
Step2247 [10]

Answer:

We have 7 complete teams

Step-by-step explanation:

Here, we have students preparing to make a team of 8 students per team. We now want to know how many complete teams they can make if they are a total of 60;

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8, 16 , 24 , 32, 40 , 48 , 56

So breaking it in 8s, we have;

8 8 8 8 8 8 8

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So there would be four left overs

3 0
3 years ago
find three consecutive even integers such that if the largest integer is subtracted from two times the smallest the result is 18
Vladimir [108]
Good question let me see if I can figure it out
4 0
3 years ago
Simplify <br> (2u^4/4uv^-4)^-3
aalyn [17]

4v⁶/ u⁵

Simplify

Simplify the expression.

4v6u54v6u5

Tap to view steps...

6 0
2 years ago
Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)
Triss [41]

g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}

The surface element is

\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta

and the integral is

\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta

=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)

###

To compute the last integral, you can integrate by parts:

u=r\implies\mathrm du=\mathrm dr

\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}

\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr

For this integral, consider a substitution of

r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds

\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds

\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds

=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds

and the result above follows.

4 0
3 years ago
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