The area of the plane region bounded by the curves x+2y^2=0 and x+3y^2=1 is equal to:

Question

The area of the plane region bounded by the curves $x+2y^2=0$ and $x+3y^2=1$ is equal to:

Accepted Answer

$x=-2y^2$ and $x=1-3y^2$ meet where $-2y^2=1-3y^2 \Rightarrow y^2=1 \Rightarrow y=\pm1$.

$$\text{Area} = \int_{-1}^1\left[(1-3y^2)-(-2y^2)\right]dy = \int_{-1}^1(1-y^2)\,dy = \left[y-\frac{y^3}{3}\right]_{-1}^1 = \frac43$$

Solutions · 1