일반 영역의 이중적분

(a) The integral with respect to $x$ treats $y$ as a constant. So

$$\begin{aligned} \int_0^3 f(x,y)\,\mathrm{d}x &= \int_0^3 12 x^2y^3\,\mathrm{d}x = \Big[4x^3y^3\Big]_{x=0}^{x=3} = 108 y^3 \end{aligned}$$

(b) The integral with respect to $y$ treats $x$ as a constant. So

$$\begin{aligned} \int_0^2 f(x,y)\,\mathrm{d}y &= \int_0^2 12 x^2y^3\,\mathrm{d}y = \Big[3x^2y^4\Big]_{y=0}^{y=2} = 48x^2 \end{aligned}$$

(c) By part (a)

$$\begin{aligned} \int_0^2\int_0^3 f(x,y)\,\mathrm{d}x\,\mathrm{d}y &=\int_0^2\left[\int_0^3 f(x,y)\,\mathrm{d}x\right]\mathrm{d}y =\int_0^2 108y^3\,\mathrm{d}y =\Big[27y^4\Big]_{y=0}^{y=2} \\ &=27\times 16 = 432 \end{aligned}$$

(d) By part (b)

$$\begin{aligned} \int_0^3\int_0^2 f(x,y)\,\mathrm{d}y\,\mathrm{d}x &=\int_0^3\left[\int_0^2 f(x,y)\,\mathrm{d}y\right]\mathrm{d}x =\int_0^3 48x^2\,\mathrm{d}y =\Big[16x^3\Big]_{x=0}^{x=3} \\ &=16\times 27 = 432 \end{aligned}$$

(e) This time

$$\begin{aligned} \int_0^3\int_0^2 f(x,y)\,\mathrm{d}x\,\mathrm{d}y &=\int_0^3\left[\int_0^2 12 x^2y^3\,\mathrm{d}x\right]\mathrm{d}y =\int_0^3\left[4 x^3y^3\right]_0^2\mathrm{d}y =\int_0^3 32 y^3\mathrm{d}y \\ & =\Big[8y^4\Big]_0^3 =8\times 81 = 648 \end{aligned}$$