Question

Find the integral of f(x,y) = x^4 + y^2 over the region bounded by y=x , y =2x
and x = 2.

Answer 1

Let's draw the integral region D bounded by y=x, y=2x, x=2.

The region D is bounded above by y=2x and below by y=x in the interval [0, 2] for x, hence D can be described as the set $\{(x, y) | 0\le x\le 2, x\le y \le 2x\}.$

$\iint_D f(x,y) dA=\iint_D f(x,y) dy dx=$

$=\intop_0^2\intop_x^{2x}(x^4+y^2)dydx=\intop_0^2(x^4y+\frac{y^3}{3}) |_x^{2x} dx=$

$=\intop_0^2 (x^4\cdot 2x+\frac{(2x)^3}{3}-x^4\cdot x-\frac{x^3}{3})dx=$

$=\intop_0^2 (2x^5+\frac{8x^3}{3}-x^5-\frac{x^3}{3})dx=\intop_0^2 (x^5+\frac{7x^3}{3})dx=$

$=(\frac{x^6}{6}+\frac{7x^4}{12})|_0^2=(\frac{2^6}{6}+\frac{7\cdot 2^4}{12})-(\frac{0^6}{6}+\frac{7\cdot 0^4}{12})=$

$=\frac{64}{6}+\frac{112}{12}=\frac{32}{3}+\frac{28}{3}=\frac{60}{3}=20.$

Answer. 20