The area of the region bounded by y^2=8x and y^2=16(3-x) is equal to:

Question

The area of the region bounded by $y^2=8x$ and $y^2=16(3-x)$ is equal to:

Accepted Answer

Intersection: $8x=16(3-x)\Rightarrow x=2,\ y=\pm4$.

$$\text{Area}=\int_{-4}^4\left[\left(3-\frac{y^2}{16}\right)-\frac{y^2}8\right]dy=2\int_0^4\left(3-\frac{3y^2}{16}\right)dy=6\left(4-\frac43\right)=16.$$

Solutions · 1