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adell [148]
3 years ago
5

-2(x+2)=-12 .............

Mathematics
1 answer:
Dima020 [189]3 years ago
6 0

Answer:

x=4

Step-by-step explanation:

-2(x+2)=-12

x+2=-12/-2

x+2=6

x=6-2

x=4

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18x + 8x^2 + 7 divided by 1 + 2x

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Use spherical coordinates. Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 81, above the xy-plane, an
Natasha_Volkova [10]

Answer:

The volume of the solid is 243\sqrt{2} \ \pi

Step-by-step explanation:

From the information given:

BY applying sphere coordinates:

0 ≤ x² + y² + z² ≤ 81

0  ≤ ρ²   ≤   81

0  ≤ ρ   ≤  9

The intersection that takes place in the sphere and the cone is:

x^2 +y^2 ( \sqrt{x^2 +y^2 })^2  = 81

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Thus; the region bounded is: 0 ≤ θ ≤ 2π

This implies that:

z = \sqrt{x^2+y^2}

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tanФ = 1

Ф = π/4

Similarly; in the X-Y plane;

z = 0

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cosФ = 0

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So here; \dfrac{\pi}{4} \leq \phi \le \dfrac{\pi}{2}

Thus, volume: V  = \iiint_E \ d V = \int \limits^{\pi/2}_{\pi/4}  \int \limits ^{2\pi}_{0} \int \limits^9_0 \rho   ^2 \ sin \phi \ d\rho \   d \theta \  d \phi

V  = \int \limits^{\pi/2}_{\pi/4} \ sin \phi  \ d \phi  \int \limits ^{2\pi}_{0} d \theta \int \limits^9_0 \rho   ^2 d\rho

V = \bigg [-cos \phi  \bigg]^{\pi/2}_{\pi/4}  \bigg [\theta  \bigg]^{2 \pi}_{0} \bigg [\dfrac{\rho^3}{3}  \bigg ]^{9}_{0}

V = [ -0+ \dfrac{1}{\sqrt{2}}][2 \pi -0] [\dfrac{9^3}{3}- 0 ]

V = 243\sqrt{2} \ \pi

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3 years ago
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Answer:

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Step-by-step explanation:

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