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xz_007 [3.2K]
3 years ago
11

A basic bead pattern ( b b b g g ) can be repeated four times to make a bracelet. Complete the blanks to represent the length of

a bracelet and show work.
4(__b +__g) = __b + __g = ___

40pts
Mathematics
1 answer:
spin [16.1K]3 years ago
7 0
4(3b+2g)=12b+8g=total beads
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Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=
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Step-by-step explanation:

Given that:

The sphere of the radius = x^2 + y^2 +z^2 = 4^2

z^2 = 16 -x^2 -y^2

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The partial derivatives of Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}

Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}

Similarly;

Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}

∴

dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA

=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA

=\sqrt{ \dfrac{16}{16-x^2-y^2}  }\ \ .dA

=\dfrac{4}{\sqrt{ 16-x^2-y^2}  }\ \ .dA

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to \sqrt{12}

dA = rdrdθ

∴

The surface area S = \int \limits _R \int \ dS

=  \int \limits _R\int  \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA

= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \  \ rdrd \theta

= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr

= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}

= 8\pi ( - \sqrt{4} + \sqrt{16})

= 8π ( -2 + 4)

= 8π(2)

= 16π

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3 years ago
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