Question

The area of the loop of the curve $${ y }^{ 2 }={ x }^{ 2 }\left( 1-x \right)$$ is ___________.

A
15/8
B
4/15
C
8/15
D
None of these

Solution

The correct option is C 8/15$$\text given: { curve }=y^{2}=x^{2}(1-x) \text { (i) }$$From (i), $$y=\pm x \sqrt{1-x}$$For, AOB(region) we have $$y=+x \sqrt{1-x}$$ (From the graph which suggests that curve is in $$1^{\text {st }}$$ quadrantSo, the desired area,Area $$=2 \int^{1}_{0} x \sqrt{1-x} d x$$take, $$1-x=t^{2}$$differentiating, $$\quad-d x=2$$ td $$t$$also, limits will changefor $$x=1$$ ; for $$x=0$$$$t=0 ; \quad t=1$$$$\text { Area } =4 \int^{0}_{1}\left(1-t^{2}\right)(-t d t)(t) \\$$$$=4 \int^{0}_{1}\left(1-t^{2}\right) + \left(- t\right) d t \\$$$$=4 \int_{0}^{1}\left(t^{2}-t^{4}\right) d t \\$$$$=4\left(\frac{t^{3}}{3}-\frac{t^{5}}{5}\right)_{0}^{1} \\$$$$=4\left(\frac{1}{3}-\frac{1}{5}\right) \quad sq . \text { units } \\$$$$=4 \times \frac{2}{15} \mathrm{Sq.} \text { units } \\$$$$\text { Area } =\frac {8}{15} \mathrm { sq.units}$$Maths

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