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keigleyhannah30
25.01.2020 •
Mathematics
Let d be the region bounded by the paraboloids; z = 6 - x² - y² and z = x² + y².
write six different triple iterated integrals for the volume of d. evaluate one of the integrals.
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Ответ:
∫∫∫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
Ответ:
Step-by-step explanation:
F(x)= 1/2 x^2