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marta [7]
3 years ago
5

Explain how you can find the missing number in 3 4/5 / missing number = 2 5/7

Mathematics
1 answer:
hram777 [196]3 years ago
3 0
Cause you have the slashes and 2/4/5/6/7and 3/4/5/6/7
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7, -7

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Give a situation where"time" is the dependent variable and "distance" is independent variable .
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Let D be the region bounded by the paraboloids; z = 6 - x² - y² and z = x² + y².
Liono4ka [1.6K]

Answer:

∫∫∫1 dV=4\sqrt{3}π

Step-by-step explanation:

From Exercise we have  

z=6-x^{2}-y^{2}

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

we get

2z=6

z=3

x^{2}+y^{2}=3

We use the polar coordinates, we get

x=r cosθ

y=r sinθ

x^{2}+y^{2}&=r^{2}

r^{2}=3

We get at the limits of the variables that well need for our integral

x^{2}+y^{2}≤z≤3

0≤r ≤\sqrt{3}

0≤θ≤2π

Therefore, we get a triple integral

\int \int \int 1\, dV&=\int \int \left(\int_{x^2+y^2}^{3} 1\, dz\right) dA

=\int \int \left(z|_{x^2+y^2}^{3} \right) dA

=\int \int\ \left(3-(x^2+y^2) \right) dA

=\int \int\ \left(3-r^2 \right) dA

=\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} (3-r^2) dr dθ

=3\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}  1 dr dθ-\int_{0}^{2\pi}\int_{0}^{\sqrt{3}} r^2 dr dθ

=3\int_{0}^{2\pi} r|_{0}^{\sqrt{3}}  dθ-\int_{0}^{2\pi} \frac{r^3}{3}|_{0}^{\sqrt{3}}dθ

=3\sqrt{3}\int_{0}^{2\pi} 1 dθ-\sqrt{3}\int_{0}^{2\pi} 1 dθ

=3\sqrt{3} ·2π-\sqrt{3}·2π

=4\sqrt{3}π

We get

∫∫∫1 dV=4\sqrt{3}π

7 0
3 years ago
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