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OLga [1]
4 years ago
5

Suppose a > 0 is constant and consider the parameteric surface sigma given by r(phi, theta) = a sin(phi) cos(theta)i + a sin(

phi) j + a cos(phi) k. 0 lessthanorequalto theta lessthanorequalto 2 pi, 0 lessthanorequalto phi lessthanorequalto pi. (a) Directly verify algebraically that r parameterizes the sphere x^2 + y^2 + z^2 = a^2, by substituting x = a sin(phi), y = a sin(phi) sin(theta), and z = a cos(phi) into the left-hand side of the equation. (b) Find r_phi, r_theta, r_phi times r_theta, and |r_phi times r_theta|. (c) Compute the surface area of the sphere doubleintegral_sigma l dS using change of variables. Find the surface area of the band sigma cut from the paraboloid z = x^2 + y^2 by the planes z = 2 and z = 6 by first finding a parameterization for the surface and then computing doubleintegral_sigma dS. Find the flux of the field F = x^2j - xzk across the surface cut by the parabolic cylinder y = x^2, -1 lessthanorequalto x lessthanorequalto 1, by the planes z = 0 and z = 2, Your normal vector should point in the direction indicated in the figure below.
Mathematics
1 answer:
Gnom [1K]4 years ago
8 0

\Sigma should have parameterization

\vec r(\varphi,\theta)=a\sin\varphi\cos\theta\,\vec\imath+a\sin\varphi\sin\theta\,\vec\jmath+a\cos\varphi\,\vec k

if it's supposed to capture the sphere of radius a centered at the origin. (\sin\theta is missing from the second component)

a. You should substitute x=a\sin\varphi\cos\theta (missing \cos\theta this time...). Then

x^2+y^2+z^2=(a\sin\varphi\cos\theta)^2+(a\sin\varphi\sin\theta)^2+(a\cos\varphi)^2

x^2+y^2+z^2=a^2\left(\sin^2\varphi\cos^2\theta+\sin^2\varphi\sin^2\theta+\cos^2\varphi\right)

x^2+y^2+z^2=a^2\left(\sin^2\varphi\left(\cos^2\theta+\sin^2\theta\right)+\cos^2\varphi\right)

x^2+y^2+z^2=a^2\left(\sin^2\varphi+\cos^2\varphi\right)

x^2+y^2+z^2=a^2

as required.

b. We have

\vec r_\varphi=a\cos\varphi\cos\theta\,\vec\imath+a\cos\varphi\sin\theta\,\vec\jmath-a\sin\varphi\,\vec k

\vec r_\theta=-a\sin\varphi\sin\theta\,\vec\imath+a\sin\varphi\cos\theta\,\vec\jmath

\vec r_\varphi\times\vec r_\theta=a^2\sin^2\varphi\cos\theta\,\vec\imath+a^2\sin^2\varphi\sin\theta\,\vec\jmath+a^2\cos\varphi\sin\varphi\,\vec k

\|\vec r_\varphi\times\vec r_\theta\|=a^2\sin\varphi

c. The surface area of \Sigma is

\displaystyle\iint_\Sigma\mathrm dS=a^2\int_0^\pi\int_0^{2\pi}\sin\varphi\,\mathrm d\theta\,\mathrm d\varphi

You don't need a substitution to compute this. The integration limits are constant, so you can separate the variables to get two integrals. You'd end up with

\displaystyle\iint_\Sigma\mathrm dS=4\pi a^2

# # #

Looks like there's an altogether different question being asked now. Parameterize \Sigma by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k

with \sqrt2\le u\le\sqrt6 and 0\le v\le2\pi. Then

\|\vec s_u\times\vec s_v\|=u\sqrt{1+4u^2}

The surface area of \Sigma is

\displaystyle\iint_\Sigma\mathrm dS=\int_0^{2\pi}\int_{\sqrt2}^{\sqrt6}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv

The integrand doesn't depend on v, so integration with respect to v contributes a factor of 2\pi. Substitute w=1+4u^2 to get \mathrm dw=8u\,\mathrm du. Then

\displaystyle\iint_\Sigma\mathrm dS=\frac\pi4\int_9^{25}\sqrt w\,\mathrm dw=\frac{49\pi}3

# # #

Looks like yet another different question. No figure was included in your post, so I'll assume the normal vector points outward from the surface, away from the origin.

Parameterize \Sigma by

\vec t(u,v)=u\,\vec\imath+u^2\,\vec\jmath+v\,\vec k

with -1\le u\le1 and 0\le v\le 2. Take the normal vector to \Sigma to be

\vec t_u\times\vec t_v=2u\,\vec\imath-\vec\jmath

Then the flux of \vec F across \Sigma is

\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=\int_0^2\int_{-1}^1(u^2\,\vec\jmath-uv\,\vec k)\cdot(2u\,\vec\imath-\vec\jmath)\,\mathrm du\,\mathrm dv

\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=-\int_0^2\int_{-1}^1u^2\,\mathrm du\,\mathrm dv

\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=-2\int_{-1}^1u^2\,\mathrm du=-\frac43

If instead the direction is toward the origin, the flux would be positive.

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Function f(z)=2z-(3+i)
Archy [21]
F(z) = 2z - (3+i)

1) second interate for z0 = i

first iterate = f(z0) = f(i) = 2i - 3 - i = i - 3 = z1
second iterate = f(z1) = f(i-3) = 2(i-3) - 3 - i = 2i - 6 - 3 - i = i - 9

2) third iterate for z0= 3 - i

first iterate = f(z0) = f(3 - i) = 2(3-i) - (3+i) = 6 -2i -3 - i =3 - 3i = z1
second iterate = f(z1) = f(3 -3i) = 2(3-3i) - (3+i) = 6 - 6i -3 -i = 3 - 7i = z2
third iterate = f(z2) = f(3 - 7i) = 2 (3-7i) - (3+i) = 6 - 14i - 3 - i = 3 -15i = z3.

3) first iterate for z0=0.5+i

first iterate = f(z0) = f(0.5 +i) = 2(0.5+i) - (3+i) = 1 +2i - 3 - i = - 2 + i = z1

4) third iterate for z0=-2-5i

first iterate = f(z0) = 2(- 2 - 5i) - (3 + i) = - 4 - 10i - 3 - i = -7 - 11i = z1
second iterate = f(z1) = 2( - 7 - 11i) - (3+i) = -14 - 22i - 3 - i = -17 -23i = z2
third iterate = f(z2) = 2(-17-23i) -(3+i) = -34 - 46i -3 - i = -37 - 47i = z3
8 0
3 years ago
PLEASE HELP
Mumz [18]
Well to do this, we need to know the least common multiple (LCM) of 6 and 8. Lets list the factors out.

6, 12, 18, 24, 30, 36
8, 16, 24, 32, 40, 48

As we can see, the least common multiple of 6 and 8 is 24.

Now we know that there are 8 plates per package and 8 goes into 24 3 times, so  naturally Shaniya needs 3 packages of plates

Hope this helped!!! :)
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3 years ago
Are these number sets the same? 1857300 : 1857800
yaroslaw [1]
We are asked to determine if the number sets 1857300 and 1857800 are the same number sets. To answer this, we need to recall and define that number sets are a collection of distinct objects such as elements and numbers. The answer to this question is that these two sets are NOT the same, the first one contains "3" while the second one contains "8".
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3 years ago
What is 16 divided by 1/4
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The answer for this equation is 4

6 0
3 years ago
Read 2 more answers
If m∠jkl = (8x – 6)° and measure of arc jml = (25x – 13)°, find measure of arc jml.
dem82 [27]

Applying the angle of intersecting secants theorem, the measure of arc JML is: 262°.

<h3>What is the Angle of Intersecting Secants Theorem?</h3>

The angle of intersecting secants theorem states that when two lines form an external angle outside a circle, the measure of the angle is half the difference between the measure of the major and minor intercepted arcs.

Thus:

m∠JKL = (measure of arc JML - measure of arc JL)/2 => angle of intersecting secants theorem

m∠JKL  = 8x - 6

measure of arc JML = 25x - 13

measure of arc JL = 360 - (25x - 13)

Plug in the values

8x - 6 = [(25x - 13) - (360 - (25x - 13))/2]

Solve for x

2(8x - 6) = [(25x - 13) - (360 - 25x + 13)]

16x - 12 = [(25x - 13) - (373 - 25x)]

16x - 12 = 25x - 13 - 373 + 25x

16x - 12 = 50x - 386

16x - 50x = 12 - 386

-34x = -374

x = 11

Measure of arc JML = 25x - 13

Plug in the value of x

Measure of arc JML = 25(11) - 13 = 262°

Learn more about angle of intersecting secants theorem on:

brainly.com/question/1626547

7 0
3 years ago
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