Question

Find the area of region between parabolas \( y^2 = 4ax \) and \( x^2 = 4ay \).

Hint:

Area between curves: Integrate difference over common limits; use appropriate variable (y here).

Answer 1

Intersection: Solve \( y^2 = 4ax \), \( x^2 = 4ay \). 
From first: \( x = \frac{y^2}{4a} \). Substitute: 
\[ \left( \frac{y^2}{4a} \right)^2 = 4a y \Rightarrow \frac{y^4}{16a^2} = 4a y \Rightarrow y^4 = 64 a^3 y \Rightarrow y^3 = 64 a^3 \Rightarrow y = 4a. \] \[ x = \frac{(4a)^2}{4a} = 4a. \] Points: (0,0), (4a, 4a). 
Area: Integrate \( \int_0^{4a} \left( x_2 - x_1 \right) dy \), where \( x_1 = \frac{y^2}{4a} \) (from \( y^2 = 4ax \)), \( x_2 = \sqrt{4a y} \) (from \( x^2 = 4ay \)). 
\[ A = \int_0^{4a} \left( \sqrt{4a y} - \frac{y^2}{4a} \right) dy = \int_0^{4a} 2\sqrt{a y} dy - \int_0^{4a} \frac{y^2}{4a} dy. \] \[ = 2\sqrt{a} \cdot \frac{2}{3} y^{3/2} \Big|_0^{4a} - \frac{1}{4a} \cdot \frac{y^3}{3} \Big|_0^{4a} = \frac{4}{3} \sqrt{a} (4a)^{3/2} - \frac{1}{12a} (4a)^3. \] \[ (4a)^{3/2} = 8 a^{3/2}, \frac{4}{3} \sqrt{a} \cdot 8 a^{3/2} = \frac{32}{3} a^2, (4a)^3 = 64 a^3, \frac{64 a^3}{12a} = \frac{16}{3} a^2. \] \[ A = \frac{32}{3} a^2 - \frac{16}{3} a^2 = \frac{16}{3} a^2. \] Answer: \( \frac{16}{3} a^2 \).