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BITSAT 2018
Exam
BITSAT
Year
2018
Subject
Mathematics
Chapter
Area Under Curves

BITSAT 2018 · Mathematics · Area Under Curves

The area of the region included between the curves x ^ 2 + y ^ 2 =...

The area of the region included between the curves x2+y2=a2and{x}^{2} + {y}^{2} = {a}^{2} and\sqrt{\left|x\right|} + \sqrt{\left|y\right|} = \sqrt{a}\left(a > 0\right) isThe area of the region included between the curves x ^ 2 + y ^ 2 = a ^ 2 and |x| + |y| = a (a > 0) is

Options

  1. A(π−23)a2\left(\pi - \dfrac{2}{3}\right){a}^{2} sq units
  2. B(23−π)a2\left(\dfrac{2}{3} - \pi\right){a}^{2} sq units
  3. C23πa2\dfrac{2}{3}\pi{a}^{2} sq units
  4. D(π+23)a2\left(\pi + \dfrac{2}{3}\right){a}^{2} sq units

Correct answer

A. (π−23)a2\left(\pi - \dfrac{2}{3}\right){a}^{2} sq units

Step-by-step solution

Given curves are

∣x∣+∣y∣=a...(i)\sqrt{\left|x\right|} + \sqrt{\left|y\right|} = \sqrt{a} . . . \left(i\right)

and x2+y2=a2...(ii){x}^{2} + {y}^{2} = {a}^{2} . . . \left(ii\right)

The area of the region included between the curves x ^ 2 + y ^ 2 = a ^ 2 and |x| + |y| = a (a > 0) is

Now, required areas =4= 4[shaded area in the first quadrant]

=4[πa24−∫0a(a−x)2dx]= 4\left[\dfrac{\pi{a}^{2}}{4} - {∫}_{0}^{a}{\left(\sqrt{a} - \sqrt{x}\right)}^{2}dx\right] [from Eqs. (i) and (ii)]

=4[πa24−∫0a(a+x−2ax)dx]= 4\left[\dfrac{\pi{a}^{2}}{4} - {∫}_{0}^{a}\left(a + x - 2\sqrt{a}\sqrt{x}\right)dx\right]

=4πa24−4[ax+x22−43ax3/2]0a= 4\dfrac{\pi{a}^{2}}{4} - 4{\left[ax + \dfrac{{x}^{2}}{2} - \dfrac{4}{3}\sqrt{a}{x}^{3 / 2}\right]}_{0}^{a}

=(π−23)a2= \left(\pi - \dfrac{2}{3}\right){a}^{2} sq units

Hence, option (a) is correct.

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