Double Integral Problem - Appreciate your help

Question

∫∫_D (2x² y) dA, where the region D is bounded by y=3x² and y=16-x²

I want to know how to represent it/ set it up in terms of both dx dy, AND dy dx.

Answer 1

3x² = 16 - x²; 4x² = 16; x = ±2; D: - 2 ≤ x ≤ 2, 3x² ≤ y ≤ 16 - x². ∫∫_D2x²ydA = ∫_-2²2x²∫_3x^2^{16-x^2}ydydx =

∫_-2²x²[(16 - x²)² - 9x⁴]dx = 2∫₀²x²(256 - 32x² - 8x⁴)dx = 2(256x³/3 - 32x⁵/5 - 8x⁷/7)₀² = 69632/105;

In terms of dxdy: {0 ≤ y ≤ 12, -√y/3 ≤ x ≤ √y/3} U {12 ≤ y ≤ 16, -√16 - y ≤ x ≤ √16 - y}.

But integration in this order need much more efforts