Question

use cylindrical coordinates. evaluate (x y z) dv e , where e is the solid in the first octant that lies under the paraboloid z = 16 − x2 − y2.

Answer 1

In converting to cylindrical coordinates, we take

x(r, θ, z) = r cos(θ)

y(r, θ, z) = r sin(θ)

z(r, θ, z) = z

so that the Jacobian under this transformation is

$J = \begin{bmatrix}x_r & x_\theta & x_z \\ y_r & y_\theta & y_z \\ z_r & z_\theta & z_z\end{bmatrix} = \begin{bmatrix}\cos(\theta) & -r\sin(\theta) & 0 \\ \sin(\theta) & r\cos(\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix}$

with det(J) = r.

In cylindrical coordinates, the region E is the set

$E = \left\{ (r, \theta, z) : 0 \le r \le 4 \text{ and } 0 \le \theta \le \dfrac\pi2 \text{ and } 0 \le z \le 16 - r^2\right\}$

I assume the function you're integrating over E is missing some plus signs. Then the integral is

$\displaystyle \iiint_E (x + y + z) \, dV = \int_0^{\pi/2} \int_0^4 \int_0^{16-r^2} (r\cos(\theta) + r\sin(\theta) + z) r \, dz \, dr \, d\theta$

$\displaystyle \cdots = \int_0^{\pi/2} \int_0^4 \left(r^2(16-r^2)\cos(\theta) + r^2(16-r^2)\sin(\theta) + \frac12 r(16-r^2)^2\right) \, dr \, d\theta$

$\displaystyle \cdots = \int_0^{\pi/2} \left(\frac{2048}{15}\cos(\theta) + \frac{2048}{15}\sin(\theta) - \frac{1280}3\right) \, d\theta$

$\displaystyle \cdots = \boxed{\frac{4096}{15} + \frac{512\pi}3}$