Question

Evaluate the double integral as iterated integral in two ways : RR

R

xy2 dA; R is the region enclosed by y = 1, y = 2, x = 0,

and y = x

Answer 1

$$\displaystyle\int_{1}^2\displaystyle\int_{0}^yxy^2dxdy=\displaystyle\int_{1}^2y^2[\dfrac{x^2}{2}]\begin{matrix} y \\ 0 \end{matrix}dy$$

$$=\displaystyle\int_{1}^2\dfrac{y^4}{2}dy=[\dfrac{y^5}{10}]\begin{matrix} 2 \\ 1 \end{matrix}=\dfrac{1}{10}(32-1)=\dfrac{31}{10}$$

$$\displaystyle\int_{0}^1\displaystyle\int_{1}^2xy^2dydx+\displaystyle\int_{1}^2\displaystyle\int_{x}^2xy^2dydx$$

$$=\displaystyle\int_{0}^1x[\dfrac{y^3}{3}]\begin{matrix} 2 \\ 1 \end{matrix}dx+\displaystyle\int_{1}^2x[\dfrac{y^3}{3}]\begin{matrix} 2 \\ x \end{matrix}dyx$$

$$=\displaystyle\int_{0}^1\dfrac{x}{3}(8-1)dx+\displaystyle\int_{1}^2\dfrac{x}{3}(8-x^3)dx$$

$$=\dfrac{7}{3}[\dfrac{x^2}{2}]\begin{matrix} 1 \\ 0 \end{matrix}+\dfrac{1}{3}[4x^2-\dfrac{x^5}{5}]\begin{matrix} 2 \\ 1 \end{matrix}$$

$$=\dfrac{7}{6}+\dfrac{1}{3}(16-\dfrac{32}{5}-(4-\dfrac{1}{5}))=\dfrac{7}{6}+\dfrac{29}{15}$$

$$=\dfrac{83}{30}$$