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Area Under the Curves question

2014 · Shift 0 · Q29
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Area Under the Curves question

2014 · Shift 0 · Q29

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area of the region described by A={(x,y):x2+y2≤1A = \left\{ {\left( {x,y} \right):{x^2} + {y^2} \le 1} \right.A={(x,y):x2+y2≤1 and y2≤1−x}\left. {{y^2} \le 1 - x} \right\}y2≤1−x} is :
  1. A
    π2−23{\pi \over 2} - {2 \over 3}2π​−32​
  2. B
    π2+23{\pi \over 2} + {2 \over 3}2π​+32​
  3. C
    π2+43{\pi \over 2} + {4 \over 3}2π​+34​
  4. D
    π2−43{\pi \over 2} - {4 \over 3}2π​−34​
View written solutionFree

Correct answer: C

  1. Interpret the region

The set is

A={(x,y):x2+y2≤1 and y2≤1−x}.A=\{(x,y): x^2+y^2\le 1 \text{ and } y^2\le 1-x\}.A={(x,y):x2+y2≤1 and y2≤1−x}.

So the region is the common part of:

  • the unit circle: x2+y2≤1,x^2+y^2\le 1,x2+y2≤1,
  • the parabola-like region: y2≤1−x  ⟺  x≤1−y2.y^2\le 1-x \iff x\le 1-y^2.y2≤1−x⟺x≤1−y2.

We need the area of their intersection.

  1. Rewrite both curves as boundaries in terms of xxx

From the circle, x2+y2=1  ⟹  x=±1−y2.x^2+y^2=1 \implies x=\pm\sqrt{1-y^2}.x2+y2=1⟹x=±1−y2​. Since the circle interior is included, −1−y2≤x≤1−y2.-\sqrt{1-y^2}\le x\le \sqrt{1-y^2}.−1−y2​≤x≤1−y2​.

From y2≤1−x,y^2\le 1-x,y2≤1−x, we get x≤1−y2.x\le 1-y^2.x≤1−y2.

Thus for a fixed yyy, the allowed xxx must satisfy −1−y2≤x≤min⁡(1−y2, 1−y2).-\sqrt{1-y^2}\le x\le \min\left(\sqrt{1-y^2},\,1-y^2\right).−1−y2​≤x≤min(1−y2​,1−y2).

Also, from the circle, we must have −1≤y≤1.-1\le y\le 1.−1≤y≤1.

  1. Compare 1−y2\sqrt{1-y^2}1−y2​ and 1−y21-y^21−y2

Let t=1−y2,t=1-y^2,t=1−y2, with 0\le t\le 1.$$ Then we compare \sqrt tandandandt.For. For .For0\le t\le 1$, t≥t.\sqrt t\ge t.t​≥t. Hence 1−y2≤1−y2.1-y^2 \le \sqrt{1-y^2}.1−y2≤1−y2​.

Therefore, min⁡(1−y2,1−y2)=1−y2.\min\left(\sqrt{1-y^2},1-y^2\right)=1-y^2.min(1−y2​,1−y2)=1−y2.

So for each y∈[−1,1]y\in[-1,1]y∈[−1,1], the horizontal width of the region is

(1−y2)−(−1−y2)=1−y2+1−y2.(1-y^2)-\left(-\sqrt{1-y^2}\right)=1-y^2+\sqrt{1-y^2}.(1−y2)−(−1−y2​)=1−y2+1−y2​.

Hence area is

Area=∫−11(1−y2+1−y2) dy.\text{Area}=\int_{-1}^{1}\left(1-y^2+\sqrt{1-y^2}\right)\,dy.Area=∫−11​(1−y2+1−y2​)dy.
  1. Evaluate the integral

Split it:

Area=∫−11(1−y2) dy+∫−111−y2 dy.\text{Area}=\int_{-1}^{1}(1-y^2)\,dy+\int_{-1}^{1}\sqrt{1-y^2}\,dy.Area=∫−11​(1−y2)dy+∫−11​1−y2​dy.

First integral

∫−11(1−y2) dy=[y−y33]−11=(1−13)−(−1+13)=23+23=43.\int_{-1}^{1}(1-y^2)\,dy =\left[y-\frac{y^3}{3}\right]_{-1}^{1} =\left(1-\frac13\right)-\left(-1+\frac13\right) =\frac23+\frac23 =\frac43.∫−11​(1−y2)dy=[y−3y3​]−11​=(1−31​)−(−1+31​)=32​+32​=34​.

Second integral

∫−111−y2 dy\int_{-1}^{1}\sqrt{1-y^2}\,dy∫−11​1−y2​dy

is the area of the upper semicircle of radius 111, hence

∫−111−y2 dy=π2.\int_{-1}^{1}\sqrt{1-y^2}\,dy=\frac{\pi}{2}.∫−11​1−y2​dy=2π​.

So,

Area=43+π2.\text{Area}=\frac43+\frac{\pi}{2}.Area=34​+2π​.
  1. Match with the options
Area=π2+43\boxed{\text{Area}=\frac{\pi}{2}+\frac{4}{3}}Area=2π​+34​​

which corresponds to Option C.

  1. Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.

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