CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

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


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

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 $$1^{\text {st }}$$ quadrant

So, 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 change
for $$x=1$$ ; 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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image