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Zarrin [17]
3 years ago
13

Given that curl F = 2yi – 2zj + 3k, find the surface integral of the normal component of curl F (not F) over (a) the open hemisp

herical surface x2 + y2 + z2 = 9, z > 0. (b) the sphere x2 + y2 + z2 = 9. (In both parts, you should be able to write the answer down by inspection.)
Mathematics
1 answer:
Dimas [21]3 years ago
7 0

Use Stokes' theorem for both parts, which equates the surface integral of the curl to the line integral along the surface's boundary.

a. The boundary of the hemisphere is the circle x^2+y^2=9 in the plane z=0, where the curl is \mathrm{curl}\vec F=2y\,\vec\imath+3\,\vec k. Green's theorem applies here, so that

\displaystyle\iint_S\mathrm{curl}\vec F\cdot\mathrm d\vec S=\int_{\partial S}\vec F\cdot\mathrm d\vec r=3\int_{x^2+y^2=9}\mathrm d\vec r

which means the value of the line integral is 3 times the area of the circle, or 27\pi.

b. The closed sphere has no boundary, so by Stokes' theorem the integral is 0.

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In a right triangle, if sin 0 = 8/17 what is cos 0 explain?
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2 years ago
Average box of crackers is 24.5 ounces with standard deviation of. 8 ounce. What percent of the boxes weigh more than 22.9 ounce
34kurt

Answer:

97.7% of of the boxes weigh more than 22.9 ounces.

15.9% of of the boxes weigh less than 23.7 ounces.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  24.5 ounces

Standard Deviation, σ = 0.8 ounce

We are given that the distribution of boxes weight is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

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P( x > 22.9) = P( z > \displaystyle\frac{22.9 - 24.5}{0.8}) = P(z > -2)

= 1 - P(z \leq -2)

Calculation the value from standard normal z table, we have,  

P(x > 22.9) = 1 - 0.023 =0.977= 97.7\%

97.7% of of the boxes weigh more than 22.9 ounces.

b) P(boxes weigh less than 23.7 ounces)

P(x < 23.7)

P( x < 23.7) = P( z < \displaystyle\frac{23.7 - 24.5}{0.8}) = P(z < -1)

Calculation the value from standard normal z table, we have,  

P(x < 23.7) =0.159= 15.9\%

15.9% of of the boxes weigh less than 23.7 ounces.

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3 years ago
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