Evaluate the double integral of 9y + 4x over the region bounded by the parabolas y = 4x² − 9 and y = 9 − 4x².

Step-by-step solution

1. 1

   Find where the two parabolas meet. Setting $4x^2-9 = 9-4x^2$ gives $8x^2 = 18$, so $x = \pm\tfrac{3}{2}$. The region therefore spans $-\tfrac{3}{2} \le x \le \tfrac{3}{2}$.

2. 2

   Write the iterated integral. For each $x$, the vertical slice runs from the lower parabola to the upper one:

   $\int_{-3/2}^{3/2}\int_{4x^2-9}^{9-4x^2}(9y+4x)\,dy\,dx$

3. 3

   Exploit the symmetry instead of grinding it out. $D$ is symmetric about both axes: swapping $x \to -x$ maps the region to itself, and so does $y \to -y$ (the two boundary curves swap).

4. 4

   Apply the symmetry to each term. $9y$ is odd in $y$ over a region symmetric in $y$, so it integrates to $0$. $4x$ is odd in $x$ over a region symmetric in $x$, so it also integrates to $0$.

5. 5

   Conclude. Both terms vanish, so the integral is $0$ — no antiderivative needed. Checking for symmetry before integrating is worth doing every time a region is defined by even functions.