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Gelneren [198K]
3 years ago
8

Find the area of the part of the sphere x2 + y2 + z2 = 100z that lies inside the paraboloid z = x2 + y2.

Mathematics
1 answer:
vekshin13 years ago
6 0
x^2+y^2+z^2=100z\iff x^2+y^2+(z-50)^2=50^2

The surface of interest can be parameterized by

\mathbf r(u,v)=\begin{cases}x(u,v)=50\cos u\sin v\\y(u,v)=50\sin u\sin v\\z(u,v)=50(\cos v+1)\end{cases}

with 0\le u\le2\pi and 0\le v\le\arccos\dfrac{49}{50}.

The area of the surface S is given by the surface integral

\displaystyle\iint_S\mathrm dA=\iint_T\left\|\mathbf r_u\times\mathbf r_v\right\|\,\mathrm du\,\mathrm dv

We have

\|\mathbf r_u\times\mathbf r_v\|=50^2\sin v

so the area is given by the double integral

\displaystyle50^2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\arccos(49/50)}\sin v\,\mathrm dv\,\mathrm du=100\pi
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Step-by-step explanation:

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The mixed number have to become a improper fraction and then you have to subtract like this.

so 7 7/8 you have to multiply 7 times 8 and then add it by 7 .
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63/8 - 26/8 = then you have to do 63 - 26 = 37 so is 37/8 and then simplify or divide .

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The width of a rectangle is 6 in. less than its length. The perimeter is 68 in. What is the width of the rectangle?
stepan [7]
Length= y in.
Width= (y-6) in.
Perimeter = 2(L*W)
Perimeter=2L+2W
68in.=2(y)+2(y-6)
68=2y+2y-12
68=4y-12
68+12=4y
4y=80
y=20
Length= 20 inches, width =14 inches
7 0
3 years ago
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