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

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

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

SOLUTION Projection of the solid bounded by two cylinders onto $xy$-plane is $x^2 + y^2 = a^2$ and $x^2 = a^2$. 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

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

Therefore the volume of the solid is

$$V = 2\iint_{\Omega}\sqrt{a^2 - x^2}dxdy = 2 \int_{-a}^{a}\big(\int_{-\sqrt{a^2 - x^2}}^{\sqrt{a^2 - x^2}} \sqrt{a^2 - x^2}dy\big)dx$$

$$= 8\int_{0}^{a}\big(\int_{y = 0}^{\sqrt{a^2 - x^2}}\sqrt{a^2 - x^2}dy\big)dx$$

$$= 8\int_{0}^{a}\big[\sqrt{a^2 - x^2}y\big]_0^{\sqrt{a^2 - x^2}}dx$$

$$= 8\int_{0}^{a}(a^2 - x^2)dx = 8\left[a^2 x - \frac{x^3}{3}\right ]_{0}^{a} = \frac{16a^3}{3} \ensuremath{ \blacksquare}$$

Exercise 5..8   Find the volume of the solid bounded by $${x^2 + y^2 \leq a^2}$$ 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 $x = r\cos{\theta} > 0$,

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

Thus,

$$V = \iint_{\Omega}x dx dy = \iint_{\Gamma}r\cos{\theta} (r) dr d\theta$$

$$= 2\int_{0}^{\frac{\pi}{2}}\int_{0}^{a}r^2 \cos{\theta} dr d\theta = 2\int_{0}^{\frac{\pi}{2}}\cos{\theta}d\theta \int_{0}^{a}r^2 dr$$

$$= 2\left[\sin{\theta}\right]_{0}^{\frac{\pi}{2}}\left[\frac{r^3}{3}\right]_{0}^{a} = \frac{2a^3}{3}\ensuremath{ \blacksquare}$$

Exercise A

1.

Find the area of the region satisfying the following conditions．

(a) The region enclosed by $${r = 1 - \cos{\theta}}$$

(b) Inside of the curve $${r = 3\cos{\theta}}$$ and outside of the curve $r = \frac{3}{2}$.

(c) Inside of the curve $${r = 3\cos{\theta}}$$ and outside of the curve $r = \cos{\theta}$

2.

Find the surface area of the following surface．

(a) $z^{2} = x^{2} + y^{2}$ with $0 \leq x^{2} + y^{2} \leq 1$, $z \geq 0$.

(b) $x+y+z = 2$ with $x \geq 0, y \geq 0, z \geq 0$.

(c) $z = xy$ with $x^{2} + y^{2} = a^{2} (a > 0)$.

3.

Find the volume of the following solid.

(a) $z = x^{2} + y^{2}$ with $z = 2y$

(b) $z = x^{2} + y^{2}$ with $x^{2} + y^{2} \leq 1$

Exercise B

1.

Find the area of the region enclosed by the following curves.

(a) $${x = \cos^{3}{t}, y = \sin^{3}{t} (0 \leq t \leq \frac{\pi}{2})}$$

(b) $${r = a\cos{3\theta} (a > 0)}$$ (c) $${y = \frac{8}{x^2 + 4}}$$ and $${y = \frac{x^2}{4}}$$.

2.

Find the surface area of the following surface．

(a) Sphere $${x^2 + y^2 + z^2 = a^2}$$ with the radius $a$.

(b) $z = xy$ with $${x^2 + y^2 \leq a^2}$$

(c) $${x^2 + z^2 = a^2}$$ and $${x^2 + y^2 = a^2}$$.

(d) The surface generated by rotating $y = mx (0 \leq x \leq k)$ about $x$-axis for $(m > 0)$.

3.

Find the volume of the solid given by the following conditions.．

(a) $${x^2 + y^2 \leq a^2}$$ with $0 \leq z \leq x$

(b) $${0 \leq z \leq 1 - x^2, x \leq 1 - y^2, x \geq 0, y \geq 0}$$.

(c) $${x^2 + y^2 + z^2 \leq a^2}$$ and $${x^2 + y^2 \leq ax}$$.

(d) $${z = 1 - \sqrt{x^2 + y^2}}$$ and $z = x$, $x = 0$.