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Evaluate the volume of the solid $$A=\{(x,y,z)\in \mathbb{R}^3: x^2+y^2-2y\le0; 0\le z\le 10-3\sqrt {x^2+y^2}\}.$$

Knowing that integration "by wires" can be used when the domain of integration $\Omega$ can be written as $$\Omega=\{(x,y,z)\in \mathbb{R}^3: g_1(x,y)\le z\le g_2(x,y); (x,y)\in D\}$$ we can setup the integral $$\begin{align}\iiint_{A}dxdydz&=\iint_{x^2+y^2-2y\le 0}\Big(\int_{0}^{10-3\sqrt{x^2+y^2}}dz\Big)dxdy \\&=\iint_{x^2+y^2-2y\le 0}\Big(10-3\sqrt {x^2+y^2}\Big)dxdy.\end{align}$$ Now, the domain $D=\{(x,y)\in \mathbb{R}^2: x^2+y^2-2y\le 0\}$ is a circle with centre in $(0,1)$ and radius $1$. I used the following polar coordinates: $x=r\cos(\theta)$, $y=1+r\sin(\theta)$ because of the translation of the circle and the integral became: $$\int_{0}^{2\pi}\int_{0}^{1}(10-3\sqrt{r^2\cos^2(\theta)+1+2r\sin(\theta)+r^2\sin^2(\theta)})rdrd\theta\\ =\int_{0}^{2\pi}\int_{0}^{1}\Big(10-3\sqrt{r^2+1+2r\sin(\theta)}\Big)rdrd\theta.$$ As the final calculations are not so straightforward I'm perplexed about my method and, in the first place, I'm asking if this is correct, in the second place if there exists another method which is simpler.

Robert Z
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  • Usually, an area is a simple integral, and a volume is a double integral. I don't see any reason to have triple integral to calculate a volume. – Lourrran Apr 14 '23 at 07:59
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    By definition given $A\subset R^2$, $Area(A)=\iint_{A}dxdy$ and given $B\subset R^3$, $Volume(B)=\iiint_{B}dxdydz$. At least, this is what I know. –  Apr 14 '23 at 08:20

1 Answers1

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Since, by definition:

$$ ||A|| := \iiint\limits_A 1\,\text{d}x\,\text{d}y\,\text{d}z, $$

$$ A := \left\{(x,y,z)\in\mathbb{R}^3 : x^2+y^2-2y \le 0, \, 0 \le z \le 10-3\sqrt{x^2+y^2}\right\} $$

it's convenient to start by integrating for wires parallel to the z-axis, from which:

$$ ||A|| = \iint\limits_B \left(10-3\sqrt{x^2+y^2}\right)\text{d}x\,\text{d}y, $$

$$ B := \left\{(x,y)\in\mathbb{R}^2 : x^2+y^2-2y \le 0, \, 0 \le 10-3\sqrt{x^2+y^2}\right\}. $$

Therefore, passing to polar coordinates in the plane with center $O$, we have:

$$ \begin{cases} \rho \ge 0 \\ 0 \le \theta < 2\pi \\ \rho^2-2\rho\sin\theta \le 0 \\ 0 \le 10-3\rho \\ \end{cases} \quad \quad \Leftrightarrow \quad \quad \begin{cases} 0 \le \rho \le 2\sin\theta \\ 0 \le \theta \le \pi \\ \end{cases} $$

it all boils down to the computation of two iterated single integrals:

$$ ||A|| = \int_0^{\pi}\left(\int_0^{2\sin\theta}\left(10-3\rho\right)\rho\,\text{d}\rho\right)\text{d}\theta = \boxed{10\pi-\frac{32}{3}}. $$

πρόσεχε
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