Name: Area Enclosed by a Curve and a Line: Illustration
Uploaded: 2019-07-19
Duration: 3 min 35 s
Description: How would you find the area enclosed between a parabola and a straight line? Watch the video to see the method as an application of definite integrals.

Question

Find the area of the figure which lies in the

1^{st}

quadrant and is bounded by the curves

x^{2} + y^{2} = 3 a^{2}, y^{2} = 2 a x

and

x^{2} = 2 a y, (a > 0) .

Accepted Answer

Given curves are $x^{2} + y^{2} = 3 a^{2}, y^{2} = 2 a x & x^{2} = 2 a y; a > 0$

Point of intersection of curves $y^{2} = 2 a x & x^{2} + y^{2} = 3 a^{2}$ is $(a, \sqrt{2} a)$

Point of intersection of curves $x^{2} = 2 a y & x^{2} + y^{2} = 3 a^{2}$ is $(\sqrt{2} a, a)$

$∴ R e q u i r e d A r e a = \int_{0}^{a} \sqrt{2 a x} d x + \int_{a}^{\sqrt{2} a} \sqrt{3 a^{2} - x^{2}} d x - \int_{0}^{\sqrt{2} a} \frac{x^{2}}{2 a} d x$

$= \sqrt{2 a} \cdot {(\frac{x^{\frac{3}{2}}}{\frac{3}{2}})}_{0}^{a} + {(\frac{x}{2} \sqrt{3 a^{2} - x^{2}} + \frac{3 a^{2}}{2} \sin^{- 1} \frac{x}{\sqrt{3} a})}_{a}^{\sqrt{2} a} - {(\frac{x^{3}}{6 a})}_{0}^{\sqrt{2} a}$

$= (\sqrt{2 a} \cdot \frac{2}{3} \cdot a^{\frac{3}{2}}) + (\frac{\sqrt{2} a}{2} \cdot a + \frac{3 a^{2}}{2} \sin^{- 1} \sqrt{\frac{2}{3}}) - (\frac{a}{2} \cdot \sqrt{2} a + \frac{3 a^{2}}{2} \sin^{- 1} \frac{1}{\sqrt{3}}) - (\frac{2 \sqrt{2} a^{3}}{6 a})$

$= \frac{2 \sqrt{2}}{3} a^{2} + \frac{a^{2}}{\sqrt{2}} + \frac{3 a^{2}}{2} (\sin^{- 1} \sqrt{\frac{2}{3}} - \sin^{- 1} \frac{1}{\sqrt{3}}) - \frac{a^{2}}{\sqrt{2}} - \frac{\sqrt{2}}{3} a^{2}$

$= \frac{\sqrt{2}}{3} a^{2} + \frac{3 a^{2}}{2} \sin^{- 1} (\sqrt{\frac{2}{3}} \sqrt{1 - \frac{1}{3}} - \frac{1}{\sqrt{3}} \sqrt{1 - \frac{2}{3}})$

$= \frac{\sqrt{2}}{3} a^{2} + \frac{3 a^{2}}{2} \sin^{- 1} \frac{1}{3} = (\frac{\sqrt{2}}{3} + \frac{3}{2} \sin^{- 1} \frac{1}{3}) a^{2} s q . u n i t$

Find the area of the figure which lies in the $1^{st}$ quadrant and is bounded by the curves $x^{2} + y^{2} = 3 a^{2}, y^{2} = 2 a x$ and $x^{2} = 2 a y, (a > 0) .$

Important Questions on Area Enclosed by Curves

Learn and Prepare for any exam you want !

Find the area of the figure which lies in the 1st quadrant and is bounded by the curves x2+y2=3a2, y2=2ax and x2=2ay, (a>0).

Important Questions on Area Enclosed by Curves

Learn and Prepare for any exam you want !

Find the area of the figure which lies in the $1^{st}$ quadrant and is bounded by the curves $x^{2} + y^{2} = 3 a^{2}, y^{2} = 2 a x$ and $x^{2} = 2 a y, (a > 0) .$