Question

Find the area of region included between the parabola y2 = 4ax and x2 = 4ay, a > 0
​

Answer 1

$\huge\boxed{\fcolorbox{red}{yellow}{Shailesh}}$

Y² = 4 a x and y = x² / 4 a

find intersection points: y² = 4 a x = x⁴ /16a²

=> x,y = (0, 0) or (4 a, 4a)

Area:

= \int\limits^a_b {[y_1-y_2]} \, dx \\\\= \int\limits^{4a}_0 {[\sqrt{4a}\sqrt{x}-x^2/4a]} \, dx \\\\=[\frac{4\sqrt{a}}{3}(x)^\frac{3}{2}-\frac{x^3}{12a}]_0^{4a}\\\\=\frac{32a^2}{3}-\frac{16a^2}{3}\\\\=\frac{16a^2}{3}

$\bf\huge\underline\red{Follow me}$