Question

Let D be the region bounded by the paraboloids; z = 6 - x² - y² and z = x² + y².

Answer 1

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