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Area Under the Curves question

2023 · 11 Apr · Shift 1 · Q36
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  5. /2023 · 11 Apr · Shift 1 · Q36

Area Under the Curves question

2023 · 11 Apr · Shift 1 · Q36

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
Area of the region {(x,y):x2+(y−2)2≤4,x2≥2y}\left\{(x, y): x^{2}+(y-2)^{2} \leq 4, x^{2} \geq 2 y\right\}{(x,y):x2+(y−2)2≤4,x2≥2y} is
  1. A
    2π+1632 \pi+\frac{16}{3}2π+316​
  2. B
    π−83\pi-\frac{8}{3}π−38​
  3. C
    π+83\pi+\frac{8}{3}π+38​
  4. D
    2π−1632 \pi-\frac{16}{3}2π−316​
View written solutionFree

Correct answer: D

  1. Interpret the region

We need the area of {(x,y):x2+(y−2)2≤4,  x2≥2y}.\{(x,y): x^2+(y-2)^2\le 4,\; x^2\ge 2y\}.{(x,y):x2+(y−2)2≤4,x2≥2y}.

This is the common region of:

  • the circle x2+(y−2)2=4,x^2+(y-2)^2=4,x2+(y−2)2=4,
  • and the region satisfying x2≥2y  ⟺  y≤x22.x^2\ge 2y \iff y\le \frac{x^2}{2}.x2≥2y⟺y≤2x2​.

So we want the part of the circle lying below the parabola y=x22.y=\frac{x^2}{2}.y=2x2​.

  1. Write the circle in upper/lower form

From x2+(y−2)2=4,x^2+(y-2)^2=4,x2+(y−2)2=4, we get (y−2)2=4−x2,(y-2)^2=4-x^2,(y−2)2=4−x2, so y=2±4−x2.y=2\pm \sqrt{4-x^2}.y=2±4−x2​.

Thus:

  • upper semicircle: y=2+4−x2,y=2+\sqrt{4-x^2},y=2+4−x2​,
  • lower semicircle: y=2−4−x2.y=2-\sqrt{4-x^2}.y=2−4−x2​.

The circle extends for −2≤x≤2.-2\le x\le 2.−2≤x≤2.

  1. Find intersection points of circle and parabola

At intersection, y=x22.y=\frac{x^2}{2}.y=2x2​. Substitute into the circle: x2+(x22−2)2=4.x^2+\left(\frac{x^2}{2}-2\right)^2=4.x2+(2x2​−2)2=4.

Expand: x2+x44−2x2+4=4,x^2+\frac{x^4}{4}-2x^2+4=4,x2+4x4​−2x2+4=4, x44−x2=0,\frac{x^4}{4}-x^2=0,4x4​−x2=0, x2(x24−1)=0.x^2\left(\frac{x^2}{4}-1\right)=0.x2(4x2​−1)=0.

Hence x=0,  ±2.x=0,\; \pm 2.x=0,±2.

Corresponding yyy values:

  • at x=0x=0x=0, y=0,y=0,y=0,
  • at x=±2x=\pm 2x=±2, y=2.y=2.y=2.

So the curves meet at (0,0),  (−2,2),  (2,2).(0,0),\; (-2,2),\; (2,2).(0,0),(−2,2),(2,2).

  1. Determine vertical limits

Inside the circle, for a fixed xxx, yyy ranges from 2−4−x2to2+4−x2.2-\sqrt{4-x^2} \quad \text{to} \quad 2+\sqrt{4-x^2}.2−4−x2​to2+4−x2​.

We additionally need y≤x22.y\le \frac{x^2}{2}.y≤2x2​.

So for each xxx, the required part is from the lower semicircle up to the parabola, provided the parabola lies inside the circle.

Between the intersection points, i.e. for −2≤x≤2,-2\le x\le 2,−2≤x≤2, the parabola lies between the lower and upper semicircles. Hence area is A=∫−22(x22−(2−4−x2))dx.A=\int_{-2}^{2}\left(\frac{x^2}{2}-\left(2-\sqrt{4-x^2}\right)\right)dx.A=∫−22​(2x2​−(2−4−x2​))dx.

So A=∫−22(x22−2+4−x2)dx.A=\int_{-2}^{2}\left(\frac{x^2}{2}-2+\sqrt{4-x^2}\right)dx.A=∫−22​(2x2​−2+4−x2​)dx.

  1. Evaluate the integral

Split it: A=∫−22x22 dx−∫−222 dx+∫−224−x2 dx.A=\int_{-2}^{2}\frac{x^2}{2}\,dx-\int_{-2}^{2}2\,dx+\int_{-2}^{2}\sqrt{4-x^2}\,dx.A=∫−22​2x2​dx−∫−22​2dx+∫−22​4−x2​dx.

(i) First integral

=\frac12\left[\frac{x^3}{3}\right]_{-2}^{2} =\frac12\cdot \frac{16}{3} =\frac{8}{3}.$$ ### (ii) Second integral $$\int_{-2}^{2}2\,dx=2(4)=8.$$ ### (iii) Third integral $$\int_{-2}^{2}\sqrt{4-x^2}\,dx$$ is the area of a semicircle of radius $2$, so $$=\frac12\pi(2)^2=2\pi.$$ Therefore, $$A=\frac{8}{3}-8+2\pi=2\pi-\frac{16}{3}.$$ --- 6. **Match with options** $$A=2\pi-\frac{16}{3}.$$ So the correct option is $$\boxed{\text{D}}.$$ --- 7. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.
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