The area of the region bounded by the curves x+3y^2=0 and x+4y^2=1 is equal to:

Question

The area of the region bounded by the curves $x+3y^2=0$ and $x+4y^2=1$ is equal to:

Accepted Answer

$x = -3y^2$ and $x = 1-4y^2$ intersect where $-3y^2 = 1-4y^2 \Rightarrow y^2=1 \Rightarrow y=\pm1$.

For fixed $y$, right curve $x=1-4y^2$, left curve $x=-3y^2$; width $= (1-4y^2)-(-3y^2) = 1-y^2$.

$$A = \int_{-1}^1(1-y^2)\,dy = \left[y-\frac{y^3}{3}\right]_{-1}^1 = \frac{4}{3}$$

Solutions · 1