PX153 - I4 - non-rectangular domain of integration

so far, the integration domain has been rectangular, but if the integration region isn't rectangular, then the following approach must be taken
domain, $R$ , is given by $y_{1} (x) \leq y \leq y_{2} (x)$ for all $a \leq x \leq b$
the function is integrated over $y$ from $y_{1} (x)$ to $y_{2 (x)}$ , and then integrated over $x$ from $a$ to $b$
can alternatively be thought as:
the function is first integrated over $x$ from $x_{1} (y)$ to $x_{2} (y)$ , and then over $y$ from $c$ to $d$
first way:

I = \iint_{R} f (x, y) . d x . d y = \int_{a}^{b} (\int_{y_{1} (x)}^{y_{2} (x)} f (x, y) . d y) . d x

eg: $y_{1} = x$ , $y_{2} = \sqrt{x}$ , and $f = x y^{2}$ . evaluate:

I = \int_{0}^{1} (\int_{y_{1} = x}^{y_{2} = \sqrt{x}} x y^{2} . d y) . d x

$$I = \int_{0}^{1} \left[ \frac{xy^{3}}{3} \right]{x}^{\sqrt{x}.dx}= \int^{1}(\frac{x^{\frac{5}{2}}}{3}- \frac{x^{4}}{3}).dx =...=\frac{1}{35}$$
- change the order of integration?
$$I= \int_{0}^{1}\left( \int_{x_{1}}^{x_{2}} xy^{2}.dx \right).dy = ... = \frac{1}{35}$$

eg: find $I = \iint_{R} (x + y)^{2} . d x . d y$ , where $R$ is the domain of integration where the area is enclosed by triangle: $(0, - 1)$ , $(1, 1)$ , $(- 1, 1)$
- first, integrate over $x$ :

\begin{aligned} I & = \int (\int_{x_{1} - \frac{y + 1}{2}}^{x_{2} = \frac{y + 1}{2}} (x + y)^{2} . d x) . d y \\ = \int_{- 1}^{1} (\frac{(\frac{3 y}{2} + \frac{1}{2})^{3}}{3} - \frac{(\frac{y}{2} - \frac{1}{2})^{3}}{3}) . d y \\ = 1 \end{aligned}

- can be verified by swapping the order of integration:

$$I = \int_{-1}^{0} \left( \int_{y_{1}=-2x-1}^{1} (x+y)^{2}.dy \right).dx + \int_{0}^{1} \left( \int_{y_{1}=2x-1}^{1} (x+y)^{2}.dy \right).dx$$
- the result must be $1$

eg: what is the volume of a sphere of radius, $a$ ?
- every point at the surface of a sphere: $x^{2} + y^{2} + z^{2} = a^{2}$
- $f (x, y) = z = \pm \sqrt{a^{2} - x^{2} - y^{2}}$
$V = 8 \int_{0}^{a} \int_{0}^{\sqrt{a^{2} - x^{2}}} \sqrt{a^{2} - x^{2} - y^{2}} . d y . d x$
- let, $y = A \sin t$ , $A = \sqrt{a^{2} - x^{2}}$ , $d y = A \cos t d t$ , $y = \sqrt{a^{2} - x^{2}} ⟹ t = \frac{π}{2}$

\begin{aligned} V & = 8 \int_{0}^{a} \int_{0}^{\frac{π}{2}} \sqrt{A^{2} - A^{2} \sin^{2} t} A \cos t . d t . d x \\ = 8 \int_{0}^{a} \int_{0}^{\frac{π}{2}} A^{2} \cos^{2} t . d t . d x \\ = 8 \int_{0}^{a} \int_{0}^{\frac{π}{2}} \frac{A^{2}}{2} (\cos 2 t + 1) . d t . d x \\ = 8 \int_{0}^{a} \frac{A^{2}}{2} [\frac{1}{2} \sin 2 t + t]_{0}^{\frac{π}{2}} . d x \\ = 8 \int_{0}^{a} \frac{A^{2}}{2} \frac{π}{2} . d x \\ = 2 π \int_{0}^{a} (a^{2} - x^{2}) . d x \\ ∴ V & = \frac{4 π}{3} a^{3} \end{aligned}