Question

Calculate the double integral ∫∫R(8x+6y+48)dA where R is the region: 0≤x≤3,0≤y≤4.

Answer 1

$\begin{aligned} \iint_R (8x + 6y + 48)\, \mathrm{d}A&= \int_0^4 \int_0^3 (8x + 6y + 48)\, \mathrm{d}x\, \mathrm{d}y\\ &=\int_0^4 (4x^2 + 6xy + 48x)\, \vert_0^3 \, \mathrm{d}y\\ &= \int_0^4 (36 + 18y + 144)\, \mathrm{d}y \\ &= \int_0^4 (180 + 18y)\, \mathrm{d}y \\&= (180y + 9y^2)\vert_0^4 \\&= 720 + 144 = 864 \end{aligned}$