Question

Find the area bounded by x= (y^2) - y and x=y-(y^2), take the elements of the area parallel to y-axis.

Answer 1

Find the area bounded by $x = y^2 - y$ and $x = y - y^2$ , take the elements of the area parallel to $y$ -axis.

$x = y^{2} - y$ - red line

$$x = \left(y - \frac {1}{2}\right) ^ {2} - \frac {1}{4}$$ $$y = \pm \sqrt {x + \frac {1}{4} + \frac {1}{2}}$$

$x = y - y^{2}$ - blue line

$$x = - \left(y - \frac {1}{2}\right) ^ {2} + \frac {1}{4}$$ $$y = \pm \sqrt {\frac {1}{4} - x} + \frac {1}{2}$$ $$y ^ {2} - y = y - y ^ {2}$$ $$y = 0 \text { and } y = 1$$ $$\begin{array}{l} S = \int_ {- \frac {1}{4}} ^ {0} \left(\left(\sqrt {x + \frac {1}{4}} + \frac {1}{2}\right) - \left(- \sqrt {x + \frac {1}{4}} + \frac {1}{2}\right)\right) d x + \int_ {0} ^ {\frac {1}{4}} \left(\left(\sqrt {\frac {1}{4} - x} + \frac {1}{2}\right) - \left(- \sqrt {\frac {1}{4} - x} + \frac {1}{2}\right)\right) d x = \\ = \int_ {- \frac {1}{4}} ^ {0} 2 \left(\sqrt {x + \frac {1}{4}} + \frac {1}{2}\right) d x + \int_ {0} ^ {\frac {1}{4}} 2 \left(\sqrt {\frac {1}{4} - x} + \frac {1}{2}\right) d x = \\ = \left(\frac {4}{3} \sqrt {\left(x + \frac {1}{4}\right) ^ {3}} + x\right) \left| - \frac {0}{4} + \left(- \frac {4}{3} \sqrt {\left(\frac {1}{4} - x\right) ^ {3}} + x\right) \right| \frac {1}{0} = \frac {4}{3} \sqrt {\left(\frac {1}{4}\right) ^ {3}} + \frac {1}{4} + \frac {1}{4} + \frac {4}{3} \sqrt {\left(\frac {1}{4}\right) ^ {3}} = \frac {5}{6} \\ \end{array}$$