202109161207 Exercises 3.5.A (Q1)

Find the volume $V$ inside the paraboloid $z=x^2+y^2$ for $0\le z\le 4$ .

Background. (Triple integral in cylindrical coordinates)

$\displaystyle{\iiint\limits_{S}f(x,y,z)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z=\iiint\limits_{S'}f(r\cos\theta ,r\sin\theta ,z)\, r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z}$ where the mapping $x=r\cos\theta$ , $y=r\sin\theta$ , $z=z$ maps the solid $S'$ in $r\theta z$ -space onto the solid $S$ in $xyz$ -space in a one-to-one manner.

_{pg.122, Chapter 3.5, Michael Corral. (2008). Vector Calculus}

Setup.

hence,

Solution.

Using vertical slices, we see that

$\begin{aligned} V & =\iint\limits_{R}(4-z)\,\mathrm{d}A \\ & = \iint\limits_{R}\big( 4-(x^2+y^2)\big)\,\mathrm{d}A\end{aligned}$

where $R=\{ (x,y):x^2+y^2\le 4\}$ is the disc in $\mathbb{R}^2$ . In polar coordinates $(r,\theta )$ we know that $x^2+y^2=r^2$ and that $R'=\{ (r,\theta ): 0\le r\le 2, 0\le\theta\le 2\pi\}$ .

Thus,

$\begin{aligned} V & = \int_{0}^{2\pi}\int_{0}^{2} (4-r^2)\, r\,\mathrm{d}r\,\mathrm{d}\theta \\ & = \int_{0}^{2\pi}\int_{0}^{2} (4r-r^3)\,\mathrm{d}r\,\mathrm{d}\theta \\ & = \int_{0}^{2\pi} \bigg[ 2r^2-\frac{r^4}{4}\bigg]\bigg|_{0}^{2}\,\mathrm{d}\theta \\ & = 4\int_{0}^{2\pi} \mathrm{d}\theta \\ & = 4(2\pi )\\ & = 8\pi\quad \textrm{(cubic units)} \end{aligned}$

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