Dated: 30-10-2024

Examples

Example

Let \(R_a\) be a region bounded by circle \(x^2 + y^2 = a^2\).
Then evaluate

\[\int_{- \infty}^{+ \infty}\int_{- \infty}^{+ \infty} e^{-(x^2 + y^2)} dx dy\]

\[= \lim_{a \to \infty} \iint_R e^{-(x^2 + y^2)} dx dy\]

\[\because e^x = \exp (x)\]

\[= \lim_{a \to \infty} \iint_R \exp\left(-(x^2 + y^2)\right) dx dy\]

\[= \lim_{a \to \infty} \int_0^{2 \pi} \int_0^a \exp\left(-r^2\right) r dr d\theta\]

\[\because \frac d {dx} a^{bx} = a^{bx} \cdot \ln (a) \cdot \frac d {dx} (bx)\]

Also applying integration by subtitution,¹ we get

\[= \lim_{a \to \infty} \int_0^{2 \pi} \left[ -\frac {\exp(-r^2)}{2} \right]_0^{a} d\theta\]

\[= \lim_{a \to \infty} \int_0^{2 \pi} \left[-\frac {\exp(-a^2)}{2} - \frac 1 2 \right] d\theta\]

\[= \lim_{a \to \infty} \frac 1 2 \left[\left(1 - \exp(-a^2) \right) \theta \right]_0^{2 \pi}\]

\[ = \pi - \lim_{a \to \infty} \frac{\pi}{\exp (-a^2)} = \pi \]

Let \(G\) be a rectangular region bounded by following

\[a \le x \le b\]

\[c \le y \le d\]

\[k \le z \le l\]

Then if \(f(x, y, z)\) is a continuous² function³ then

\[ \iiint_{G} f(x, y, z) dV = \int_{a}^{b} \int_{c}^{d} \int_{k}^{\ell} f(x, y, z) dz \, dy \, dx \]

And the order of integrals⁴ can be changed.

Evaluate

\[\iiint_S xyz \, dx dy dz\]

Where

\[S = \{(x, y, z) : x^2 + y^2 + z^2 \le 1, x \ge 0, y \ge 0, z \ge 0\}\]

\[\int_0^1\int_0^{\sqrt{1 - x^2}}\int_0^{\sqrt{1 - x^2 - y^2}} xyz \, dzdydx\]

\[= \int_0^1\int_0^{\sqrt{1 - x^2}}xy\left[\frac {z^2} 2\right]_0^{\sqrt{1 - x^2 - y^2}} \, dydx\]

Continue evaluating and we will get

\[= \frac 1 {48}\]