Question

evaluate the double integral ∬d(x2 8y)da, where d is bounded by y=x, y=x3, and x≥0.

Answer 1

D is the region

$D = \left\{(x, y) : 0 \le x \le 1 \text{ and } x^3 \le y \le x \right\}$

Then the integral would be

$\displaystyle \iint_D (x^2+8y) \, dA = \int_0^1 \int_{x^3}^x (x^2+8y) \, dy \, dx$

$\displaystyle \iint_D (x^2+8y) \, dA = \int_0^1 (x^2y+4y^2)\bigg|_{x^3}^x \, dx$

$\displaystyle \iint_D (x^2+8y) \, dA = \int_0^1 (x^3+4x^2 - x^5-4x^6) \, dx$

$\displaystyle \iint_D (x^2+8y) \, dA = \left(\frac14 x^4+\frac43 x^3 - \frac16 x^6-\frac47 x^7\right)\bigg|_0^1$

$\displaystyle \iint_D (x^2+8y) \, dA = \frac14+\frac43 - \frac16 -\frac47 = \boxed{\frac{71}{84}}$