Calculate the area of the region bounded by the parabolas y² = x and x² = y.

Open in App

Solution

$y^{2} = x is a parabola, opening sideways, with vertex at O (0, 0) and + ve x - axis as axis of symmetry x^{2} = y is a parabola, opening upwards, with vertex at O (0, 0) and + ve y - axis as axis of symmetry Soving the above two equations, x^{2} = y^{4} = y \Rightarrow y^{4} - y = 0 \Rightarrow y = 0 or y = 1 . So, x = 0 or x = 1 \Rightarrow O (0, 0) and A (1, 1) are points of intersection of two curves Consider a vertical strip of length = |y_{2} - y_{1}| and width = d x \Rightarrow Area of approximating rectangle = |y_{2} - y_{1}| d x Approximating rectangle moves from x = 0 to x = 1 \Rightarrow Area of the shaded region = \int_{0}^{1} |y_{2} - y_{1}| d x \Rightarrow A = \int_{0}^{1} (y_{2} - y_{1}) d x [As, y_{2} - y_{1} > 0 \Rightarrow |y_{2} - y_{1}| = y_{1}] \Rightarrow A = \int_{0}^{1} (\sqrt{x} - x^{2}) d x \Rightarrow A = {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^{3}}{3}]}_{0}^{1} \Rightarrow A = [\frac{2}{3} \times 1^{\frac{3}{2}} - \frac{1^{3}}{3} - 0] \Rightarrow A = \frac{2}{3} - \frac{1}{3} \Rightarrow A = \frac{1}{3} sq . units Thus, area enclosed by the curves = \frac{1}{3} sq . units$

Suggest Corrections

Similar questions

y = x^{2}

x = y^{2}

y = x^{2}

y^{2} = x

y^{2} = 4 x

x^{2} = 4 y

Find the area of the region bounded by the parabola $y^{2} = 2 p x$ , and $x^{2} = 2 p y$ .

Join BYJU'S Learning Program