Given the paraboloid $z=4x^2 + 4y^2$, the plane $z=4$, and the density function $\sigma(x, y, z) = z + 2y$, find the mass of the region bounded by the paraboloid under the plane.
Intuitively, this makes the most sense using cylindrical coordinates. And since mass is just volume multipied by density (?), here's my (incorrect) attempt:
\begin{align}
z = & \; 4x^2 + 4y^2 \implies z = 4r^2 \\
\implies& \int_0^{2\pi}\int_0^1\int^4_{4r^2} \sigma(x, y, z) \, \cdot \, rdzdrd\theta = \int_0^{2\pi}\int_0^1\int^4_{4r^2} (4r^3+2r^2\sin\theta) \; dzdrd\theta \\
= \ & \int_0^{2\pi}\int_0^1 (4r^3+2r^2\sin\theta)(4-4r^2) \; drd\theta \\
= \ & 4\int_0^{2\pi}\int_0^1 (4r^3 + 2r^2\sin\theta – 4r^5 – 2r^4\sin\theta) \; drd\theta \\
= \ & 4\int_0^{2\pi} (\frac{1}{3} + \frac{2\sin\theta}{3} – \frac{2\sin\theta}{5}) \; d\theta = 4(2\pi/3) = 8\pi/3\\
\end{align}
However, the correct answer is double of that, $16\pi/3$.
Best Answer
You almost got it right. It should be$$\int_0^{2\pi}\int_0^1\int_{4r^2}^4\bigl(\color{red}{zr}+2r^2\sin(\theta)\bigr)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\frac{16\pi}3,$$since $\sigma(r\cos(\theta),r\sin(\theta),z)=z+2r\sin(\theta)$.