Question

Evaluate \(\iiint_{E}{z}dV\)where E lies above the paraboloid \(z = {x^2} + {y^2}\) and below the plane \(z = 2 y\). Use either the Table of Integrals (on Reference Pages 6-10) or a computer algebra system to evaluate the integral.

Short answer

The required integral is \(I = \frac{8}{3} * \frac{{5\pi }}{{16}} = \frac{{5\pi }}{6}\)

Step-by-step solution

Define spherical and cylindrical coordinates.

Any of the spatial coordinates acquired by building a polar coordinate system in a plane and a linear coordinate system on a line perpendicular to the plane.

three coordinates in three-dimensional space that define the location of a point The length of its radius vector r, the angle between the vertical plane containing this vector and the x-axis, and the angle between this vector and the horizontal x–y plane are the three parameters.

Now find the volume of solid

The cylindrical coordinates \(\begin{aligned}x &= r\cos \theta \\y &= r\sin \theta \\z &= z\end{aligned}\)

The limits are \({r^2} \le z \le 2r\sin \theta \;\;\;0 \le r \le 2\sin \theta \)

\(\begin{aligned}I &= \iiint_E zdV = \int_0^\pi {\int_0^{2\sin \theta } {\int_{{\tau ^2}}^{2r\sin \theta } z } } rdrdzd\theta \\ &= \int_0^\pi {\int_0^{2\sin \theta } {\left( {\frac{{{z^2}}}{2}} \right)_{{\tau ^2}}^{2r\sin \theta }} } rdrd\theta \\ &= \frac{1}{2}\int_0^\pi {\int_0^{2\sin \theta } {\left( {4{r^2}{{\sin }^2}\theta - {r^4}} \right)} } rdrd\theta \\&= \frac{1}{2}\int_0^\pi {\int_0^{2\sin \theta } {\left( {4{r^3}{{\sin }^2}\theta - {r^5}} \right)} } drd\theta \\&= \frac{1}{2}\int_0^\pi {\left( {{r^4}{{\sin }^2}\theta - \frac{{{r^6}}}{6}} \right)_0^{2\sin \theta }} d\theta \\ &= \frac{8}{3}\int_0^\pi {{{\sin }^6}} \theta d\theta \\ \end{aligned}\)

Now evaluate \({I_1} = \int_0^\pi  {si{n^6}} \theta d\theta \)

Again

\(\begin{aligned}{\sin ^6}\theta &= {\left( {{{\sin }^2}\theta } \right)^2} * {\sin ^2}\theta \\ &= {\left( {\frac{{1 - \cos 2\theta }}{2}} \right)^2} * \left( {\frac{{1 - \cos 2\theta }}{2}} \right)\\ &= \frac{1}{8}\left( {1 - 2\cos 2\theta + \frac{{1 + \cos 4\theta }}{2}} \right)(1 - 2\cos 2\theta )\\ &= \frac{1}{{32}}(10 - 15\cos 2\theta + 6\cos 4\theta - \cos 6\theta )\\{I_1} &= \int_0^\pi {{{\sin }^6}} \theta d\theta \\ &= \frac{1}{{32}}\int_0^\pi {(10 - 15\cos 2\theta + 6\cos 4\theta - \cos 6\theta )} d\theta \\ &= \frac{1}{{32}}\left( {10\theta - \frac{{15\sin 2\theta }}{2} + \frac{{3\sin 4\theta }}{2} - \frac{{\sin 6\theta }}{6}} \right)_0^\pi \\ &= \frac{{5\pi }}{{16}}\end{aligned}\)

Therefore, the required integral is \(I = \frac{8}{3} * \frac{{5\pi }}{{16}} = \frac{{5\pi }}{6}\)