Area Under Curves

Question

The area bounded by the parabolas $y^2 = 4x$ and $x^2 = 4y$ is:

RankGuru · Accepted Answer

1. Standard Result:
The area between $y^2 = 4ax$ and $x^2 = 4by$ is given by $\frac{16ab}{3}$.
Here $4a = 4 \implies a = 1$, and $4b = 4 \implies b = 1$.
Area $= 16(1)(1)/3 = 16/3$.

2. Alternative Integration:
Intersections at $(0,0)$ and $(4,4)$. Upper curve is $y = \sqrt{4x} = 2\sqrt{x}$. Lower is $y = x^2/4$.
$A = \int_0^4 (2\sqrt{x} - \frac{x^2}{4}) dx$
$= [\frac{4x^{3/2}}{3} - \frac{x^3}{12}]_0^4$
$= (\frac{4(8)}{3} - \frac{64}{12})$
$= \frac{32}{3} - \frac{16}{3} = \frac{16}{3}$.

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