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Arturiano [62]
3 years ago
11

There it is. Help me pls

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
6 0

papi frio

Step-by-step explanation:

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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
GenaCL600 [577]

Close off the hemisphere S by attaching to it the disk D of radius 3 centered at the origin in the plane z=0. By the divergence theorem, we have

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

where R is the interior of the joined surfaces S\cup D.

Compute the divergence of \vec F:

\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2

Compute the integral of the divergence over R. Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

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

So the volume integral is

\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5

From this we need to subtract the contribution of

\displaystyle\iint_D\vec F(x,y,z)\cdot\mathrm d\vec S

that is, the integral of \vec F over the disk, oriented downward. Since z=0 in D, we have

\vec F(x,y,0)=\dfrac{y^3}3\,\vec\jmath+y^2\,\vec k

Parameterize D by

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

where 0\le u\le 3 and 0\le v\le2\pi. Take the normal vector to be

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

Then taking the dot product of \vec F with the normal vector gives

\vec F(x(u,v),y(u,v),0)\cdot(-u\,\vec k)=-y(u,v)^2u=-u^3\sin^2v

So the contribution of integrating \vec F over D is

\displaystyle\int_0^{2\pi}\int_0^3-u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac{81\pi}4

and the value of the integral we want is

(integral of divergence of <em>F</em>) - (integral over <em>D</em>) = integral over <em>S</em>

==>  486π/5 - (-81π/4) = 2349π/20

5 0
3 years ago
How to do question 22?
Aleks04 [339]

Answer:

A = 20sinθ(6 + 5 cosθ)  cm²

Step-by-step explanation:

Drop perpendiculars DE and CF to AB.

Then, we have congruent triangles ADE and BCF, plus the rectangle CDEF.

The formula for the area of the trapezium is

A = ½(a + b)h

DE = 10sinθ

AE = 10cosθ

BF = 10cosθ

EF = CD = 12 cm

AB = AE + EF  + BF = 10cosθ + 12 + 10 cosθ = 12 + 20cosθ

A = ½(a + b)h

   = ½(12 +12 + 20 cosθ) × 10 sinθ

   =(24 + 20 cosθ) × 5 sinθ

   = 4(6 + 5cosθ) × 5sinθ

   = 20sinθ(6 + 5 cosθ)  cm²

6 0
3 years ago
Please help each equation is a number from 1 to 6
Angelina_Jolie [31]
The answers r 1, 2 and 4
5 0
3 years ago
Which of the following is the most precise measurement? 13 in. 13.615 in. 13.6 in. 13.62 in.
Anna35 [415]

Answer:

13.615

Step-by-step explanation:

6 0
3 years ago
The graph 4x^2-4x-1 is shown. Use the grpah to find the estimates for the solutions of 4x^2-4x-1=0 and 4x^2 - 4x-1=2
Darina [25.2K]

Answer:

a) The estimates for the solutions of 4\cdot x^{2}-4\cdot x -1 = 0 are x_{1}\approx -0.25 and x_{2} \approx 1.25.

b) The estimates for the solutions of 4\cdot x^{2}-4\cdot x -1 = 2 are x_{1}\approx -0.5 and x_{2} \approx 1.5

Step-by-step explanation:

From image we get a graphical representation of the second-order polynomial y = 4\cdot x^{2}-4\cdot x -1, where x is related to the horizontal axis of the Cartesian plane, whereas y is related to the vertical axis of this plane. Now we proceed to estimate the solutions for each case:

a) 4\cdot x^{2}-4\cdot x -1 = 0

There are two approximate solutions according to the graph, which are marked by red circles in the image attached below:

x_{1}\approx -0.25, x_{2} \approx 1.25

b) 4\cdot x^{2}-4\cdot x -1 = 2

There are two approximate solutions according to the graph, which are marked by red circles in the image attached below:

x_{1}\approx -0.5, x_{2} \approx 1.5

5 0
3 years ago
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