Question

Find the area included between the parabolas y² = 4ax and x² = 4by.

Answer 1

To find the point of intersection of the parabolas substitute \[y = \frac{x^2}{4b}\] in \[y^2 = 4ax\]  we get 
\[\frac{x^4}{16 b^2} = 4ax\]
\[ \Rightarrow x^4 - 64a b^2 x = 0\]
\[ \Rightarrow x\left( x^3 - 64a b^2 \right) = 0\]
\[ \Rightarrow x = 0\text{ and }x = 4\sqrt[3]{a b^2}\]
\[ \Rightarrow y = 0\text{ and }y = 4\sqrt[3]{a^2 b}\]
Therefore, the required area ABCD = \[\int_0^{4\sqrt[3]{a b^2}} \left( y_1 - y_2 \right) d x\] where  \[y_1 = 2\sqrt{ax}\] and  \[y_2 = \frac{x^2}{4b}\] 
Required area = \[\int_0^{4\sqrt[3]{a b^2}} \left( y_1 - y_2 \right) d x\]
\[ = \int_0^{4\sqrt[3]{a b^2}} \left( 2\sqrt{ax} - \frac{x^2}{4b} \right) d x\]
\[ = \left[ \frac{4\sqrt{a}}{3} x^\frac{3}{2} - \frac{x^3}{12b} \right]_0^{4\sqrt[3]{a b^2}} \]
\[ = \left[ \frac{4\sqrt{a}}{3} \left( 4\sqrt[3]{a b^2} \right)^\frac{3}{2} - \frac{\left( 4\sqrt[3]{a b^2} \right)^3}{12b} \right] - \left[ \frac{4\sqrt{a}}{3} \left( 0 \right)^\frac{3}{2} - \frac{\left( 0 \right)^3}{12a} \right]\]
\[ = \frac{16ab}{3}\text{ square units }\]