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Volume of Solid

Understanding
$f(x,y) - g(x,y)$ represents the height of the rectangular solid and $\Delta x \Delta y$ represents the base area. Thus $\Delta V = (f(x,y) - g(x,y))\Delta x \Delta y$.

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 $x^2 + y^2 = a^2$ 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

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

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

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Take a point in $\Omega$ and evaluate the value of $z$ to find out which surfaces upper or lower surface.

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

Exercise5-8
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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,
$\displaystyle V$ $\displaystyle =$ $\displaystyle \iint_{\Omega}x dx dy = \iint_{\Gamma}r\cos{\theta} (r) dr d\theta$  
  $\displaystyle =$ $\displaystyle 2\int_{0}^{\frac{\pi}{2}}\int_{0}^{a}r^2 \cos{\theta} dr d\theta = 2\int_{0}^{\frac{\pi}{2}}\cos{\theta}d\theta \int_{0}^{a}r^2 dr$  
  $\displaystyle =$ $\displaystyle 2\left[\sin{\theta}\right]_{0}^{\frac{\pi}{2}}\left[\frac{r^3}{3}\right]_{0}^{a} = \frac{2a^3}{3}\ensuremath{ \blacksquare}$  

The limit of integration is constant, we can express the double integral as the product of two single integral.

Exercise A


1.
Find the area of the region satisfying the following conditions．

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

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

(c) Inside of the curve $\displaystyle{r = 3\cos{\theta}}$ 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) $\displaystyle{x = \cos^{3}{t}, y = \sin^{3}{t} (0 \leq t \leq \frac{\pi}{2})}$

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

2.
Find the surface area of the following surface．

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

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

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

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

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

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

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

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Next: Triple Integrals Up: Application of Double Integrals Previous: Surface Area   Contents   Index