Volume of Solid

Given a closed bounded region $\Omega$ and continuous functions $z = f(x,y), z = g(x,y)$ on $\Omega$ , and suppose that $f(x,y) \geq g(x,y)$ on $\Omega$ . Then the volume of solid bounded by the lines parallel to $z$ -axis through the boundary $\partial \Omega$ and the surfaces $f,g$ is given by

$\displaystyle V = \iint_{\Omega}[f(x,y) - g(x,y)]dxdy$

Example 5..8 Find the volume of the solid common to the intersecting cylinders $\displaystyle{x^2 + y^2 = a^2, x^2 + z^2 = a^2 (a > 0)}$ .

SOLUTION Projection of the solid bounded by two cylinders onto $xy$

-plane is

and

. Thus, $\Omega = \{(x,y):x^2 + y^2 \leq a^2\}$ . Also $x^2 + z^2 = a^2$

implies $z = \pm \sqrt{a^2 - x^2}$ . Thus the upper surface is $\sqrt{a^2 -x^2}$ and the lower surface is $\sqrt{a^2 -x^2}$ over $\Omega$ . From this, the volume of the solid is set up by $\sum\sum 2\sqrt{a^2 - x^2}\Delta x \Delta y$ , where $\Delta x \Delta y$ represents the small rectangle of base area and $2\sqrt{a^2 - x^2}$ represents the height of a small solid cylinder. Now use vertically simple region for $\Omega$ . Then

$\displaystyle \Omega = \{(x,y): -a \leq x \leq a, -\sqrt{a^2 -x^2} \leq y \leq \sqrt{a^2 - x^2}\}$

Therefore the volume of the solid is

$\displaystyle V$	$\displaystyle =$	$\displaystyle 2\iint_{\Omega}\sqrt{a^2 - x^2}dxdy = 2 \int_{-a}^{a}\big(\int_{-\sqrt{a^2 - x^2}}^{\sqrt{a^2 - x^2}} \sqrt{a^2 - x^2}dy\big)dx$
	$\displaystyle =$	$\displaystyle 8\int_{0}^{a}\big(\int_{y = 0}^{\sqrt{a^2 - x^2}}\sqrt{a^2 - x^2}dy\big)dx$
	$\displaystyle =$	$\displaystyle 8\int_{0}^{a}\big[\sqrt{a^2 - x^2}y\big]_0^{\sqrt{a^2 - x^2}}dx$
	$\displaystyle =$	$\displaystyle 8\int_{0}^{a}(a^2 - x^2)dx = 8\left[a^2 x - \frac{x^3}{3}\right ]_{0}^{a} = \frac{16a^3}{3} \ensuremath{ \blacksquare}$

Exercise 5..8 Find the volume of the solid bounded by $\displaystyle{x^2 + y^2 \leq a^2}$ and $0 \leq z \leq x$ .

SOLUTION Note that the solid is bounded by the surface goes through the boundary of $\Omega = \{(x,y) : x^2 + y^2 \leq a^2, x \geq 0\}$ and parallel to the $z$ -axis, and the plane $z = 0$ , $z = x$ . Using the polar coordinate to express $\Omega$ . Since $x = r\cos{\theta} > 0$ ,

$\displaystyle \Gamma = \{(r,\theta) : -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}, 0 \leq r \leq a\}$

Thus,

$\displaystyle V$	$\displaystyle =$	$\displaystyle \iint_{\Omega}x dx dy = \iint_{\Gamma}r\cos{\theta} (r) dr d\theta$
	$\displaystyle =$	$\displaystyle 2\int_{0}^{\frac{\pi}{2}}\int_{0}^{a}r^2 \cos{\theta} dr d\theta = 2\int_{0}^{\frac{\pi}{2}}\cos{\theta}d\theta \int_{0}^{a}r^2 dr$
	$\displaystyle =$	$\displaystyle 2\left[\sin{\theta}\right]_{0}^{\frac{\pi}{2}}\left[\frac{r^3}{3}\right]_{0}^{a} = \frac{2a^3}{3}\ensuremath{ \blacksquare}$