Respuesta :
The resulted integral is [tex]I=\frac{8}{3} \times \frac{5 \pi}{16}=\frac{5 \pi}{6}[/tex].
What is integrals?
In mathematics, an integral is either a number representing the region under a function's graph for a certain interval or just an added to the initial, the derivative of which is initial function (indefinite integral).
Computation of the integrals:
Step 1: We employ the equations in cylindrical coordinates.
[tex]x=r \cos \theta, y=r \sin \theta, z=z[/tex]
Thus, in cylindrical coordinate system,
E lies above the paraboloid [tex]z=r^{2}[/tex] and below the plane [tex]z=2 r \sin \theta[/tex] .
Therefore, the top part E is [tex]z=2 r \sin \theta[/tex] is the cross-section between paraboloid and the plane.
Now, at the cross-section use, [tex]r^{2}=2 r \sin \theta[/tex] and [tex]z=2 r \sin \theta[/tex] .
Thus, the limits are given as ;
[tex]r^{2} \leq z \leq 2 r \sin \theta \quad 0 \leq r \leq 2 \sin \theta[/tex]
Apply the limits as compute the integration;
[tex]\begin{aligned}I=\iiint_{E} z d V &=\int_{0}^{\pi} \int_{0}^{2 \sin \theta} \int_{\tau^{2}}^{2 r \sin \theta} z r d r d z d \theta \\&=\int_{0}^{\pi} \int_{0}^{2 \sin \theta}\left[\frac{z^{2}}{2}\right]_{r^{2}}^{2 r \sin \theta} r d r d \theta \\&=\frac{1}{2} \int_{0}^{\pi} \int_{0}^{2 \sin \theta}\left[4 r^{2} \sin ^{2} \theta-r^{4}\right] r d r d \theta\end{aligned}[/tex]
[tex]\begin{aligned}&=\frac{1}{2} \int_{0}^{\pi} \int_{0}^{2 \sin \theta}\left[4 r^{3} \sin ^{2} \theta-r^{5}\right] d r d \theta \\&=\frac{1}{2} \int_{0}^{\pi}\left[r^{4} \sin ^{2} \theta-\frac{r^{6}}{6}\right]_{0}^{2 \sin \theta} d \theta \\&=\frac{8}{3} \int_{0}^{\pi} \sin ^{6} \theta d \theta\end{aligned}[/tex]
Step 2: Now, calculate for the [tex]I_{1}=\int_{0}^{\pi} \sin ^{6} \theta d \theta[/tex].
[tex]\begin{aligned}\sin ^{6} \theta &=\left(\sin ^{2} \theta\right)^{2} \times \sin ^{2} \theta \\&=\left[\frac{1-\cos 2 \theta}{2}\right]^{2} \times\left[\frac{1-\cos 2 \theta}{2}\right] \\&=\frac{1}{8}\left(1-2 \cos 2 \theta+\cos ^{2} 2 \theta\right)(1-2 \cos 2 \theta) \\&=\frac{1}{8}\left(1-2 \cos 2 \theta+\frac{1+\cos 4 \theta}{2}\right)(1-2 \cos 2 \theta)\end{aligned}[/tex]
[tex]\begin{aligned}&=\frac{1}{16}(3-4 \cos 2 \theta+\cos 4 \theta)(1-2 \cos 2 \theta) \\&=\frac{1}{32}(10-15 \cos 2 \theta+6 \cos 4 \theta-\cos 6 \theta)\end{aligned}[/tex]
Further compute the value of
[tex]\begin{aligned}I_{1} &=\int_{0}^{\pi} \sin ^{6} \theta d \theta \\&=\frac{1}{32} \int_{0}^{\pi}(10-15 \cos 2 \theta+6 \cos 4 \theta-\cos 6 \theta) d \theta \\&=\frac{1}{32}\left[10 \theta-\frac{15 \sin 2 \theta}{2}+\frac{3 \sin 4 \theta}{2}-\frac{\sin 6 \theta}{6}\right]_{0}^{\pi} \\&=\frac{5 \pi}{16}\end{aligned}[/tex]
Therefore, the obtained integral is [tex]I=\frac{8}{3} \times \frac{5 \pi}{16}=\frac{5 \pi}{6}[/tex].
Learn more about the indefinite integrals, here
https://brainly.com/question/22008756
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The complete question is -
Use cylindrical or spherical coordinates, whichever seems more appropriate. Evaluate ∫∫∫E z dV, where E lies above the paraboloid
z = x² + y²
and below the plane z = 2y. Use either the Table of Integrals or a computer algebra system to evaluate the integral.
