The area of the region enclosed by y \leq 4 x^2, x^2 \leq 9 y and y \leq 4 , is equal to :

Area Under the Curves PYQ — JEE Main 2022 · 27 Jul · Shift 2 · Q28 with Solution

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Correct answer: D

Interpret the inequalities

We are given $y \le 4x^2, \qquad x^2 \le 9y, \qquad y \le 4.$

Rewrite the second inequality: $x^2 \le 9y \implies y \ge \frac{x^2}{9}.$

So the region satisfies $\frac{x^2}{9} \le y \le 4x^2, \qquad y \le 4.$

Thus, for each fixed $x$ , the upper boundary is the smaller of $4x^2$ and $4$ .

Find the key intersection points

(i) Intersection of $y=4x^2$ and $y=4$

$4x^2=4 \implies x^2=1 \implies x=\pm 1.$

(ii) Intersection of $y=\frac{x^2}{9}$ and $y=4$

$\frac{x^2}{9}=4 \implies x^2=36 \implies x=\pm 6.$

The parabola $y=4x^2$ meets the horizontal line $y=4$ at $x=\pm1$ , so:

for $|x|\le 1$ , upper curve is $y=4x^2$ ,
for $1\le |x|\le 6$ , upper curve is $y=4$ .

The lower curve remains $y=\frac{x^2}{9}.$

The figure is symmetric about the $y$ -axis.

Set up the area integral

Using symmetry, $\text{Area}=2\left[\int_0^1 \left(4x^2-\frac{x^2}{9}\right)dx + \int_1^6 \left(4-\frac{x^2}{9}\right)dx\right].$

Simplify the first integrand: $4x^2-\frac{x^2}{9}=\frac{35x^2}{9}.$

So $\text{Area}=2\left[\int_0^1 \frac{35x^2}{9}dx + \int_1^6 \left(4-\frac{x^2}{9}\right)dx\right].$

Evaluate the integrals

First integral

$\int_0^1 \frac{35x^2}{9}dx=\frac{35}{9}\cdot \frac{1}{3}=\frac{35}{27}.$

Second integral

=\left[4x-\frac{x^3}{27}\right]_1^6.$$ At $x=6$: $$4(6)-\frac{6^3}{27}=24-8=16.$$ At $x=1$: $$4(1)-\frac{1}{27}=4-\frac{1}{27}=\frac{107}{27}.$$ Hence, $$\int_1^6 \left(4-\frac{x^2}{9}\right)dx=16-\frac{107}{27}=\frac{432-107}{27}=\frac{325}{27}.$$ So inside the bracket, $$\frac{35}{27}+\frac{325}{27}=\frac{360}{27}=\frac{40}{3}.$$ Therefore, $$\text{Area}=2\cdot \frac{40}{3}=\frac{80}{3}.$$ --- 5. **Match with the options** $$\boxed{\frac{80}{3}}$$ So the correct option is **D**. --- 6. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.

More from Area Under the Curves

View written solutionFree

Correct answer: D

Interpret the inequalities

We are given $y \le 4x^2, \qquad x^2 \le 9y, \qquad y \le 4.$

Rewrite the second inequality: $x^2 \le 9y \implies y \ge \frac{x^2}{9}.$

So the region satisfies $\frac{x^2}{9} \le y \le 4x^2, \qquad y \le 4.$

Thus, for each fixed $x$ , the upper boundary is the smaller of $4x^2$ and $4$ .

Find the key intersection points

(i) Intersection of $y=4x^2$ and $y=4$

$4x^2=4 \implies x^2=1 \implies x=\pm 1.$

(ii) Intersection of $y=\frac{x^2}{9}$ and $y=4$

$\frac{x^2}{9}=4 \implies x^2=36 \implies x=\pm 6.$

The parabola $y=4x^2$ meets the horizontal line $y=4$ at $x=\pm1$ , so:

for $|x|\le 1$ , upper curve is $y=4x^2$ ,
for $1\le |x|\le 6$ , upper curve is $y=4$ .

The lower curve remains $y=\frac{x^2}{9}.$

The figure is symmetric about the $y$ -axis.

Set up the area integral

Using symmetry, $\text{Area}=2\left[\int_0^1 \left(4x^2-\frac{x^2}{9}\right)dx + \int_1^6 \left(4-\frac{x^2}{9}\right)dx\right].$

Simplify the first integrand: $4x^2-\frac{x^2}{9}=\frac{35x^2}{9}.$

So $\text{Area}=2\left[\int_0^1 \frac{35x^2}{9}dx + \int_1^6 \left(4-\frac{x^2}{9}\right)dx\right].$

Evaluate the integrals

First integral

$\int_0^1 \frac{35x^2}{9}dx=\frac{35}{9}\cdot \frac{1}{3}=\frac{35}{27}.$

Second integral

=\left[4x-\frac{x^3}{27}\right]_1^6.$$ At $x=6$: $$4(6)-\frac{6^3}{27}=24-8=16.$$ At $x=1$: $$4(1)-\frac{1}{27}=4-\frac{1}{27}=\frac{107}{27}.$$ Hence, $$\int_1^6 \left(4-\frac{x^2}{9}\right)dx=16-\frac{107}{27}=\frac{432-107}{27}=\frac{325}{27}.$$ So inside the bracket, $$\frac{35}{27}+\frac{325}{27}=\frac{360}{27}=\frac{40}{3}.$$ Therefore, $$\text{Area}=2\cdot \frac{40}{3}=\frac{80}{3}.$$ --- 5. **Match with the options** $$\boxed{\frac{80}{3}}$$ So the correct option is **D**. --- 6. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.

Area Under the Curves question

Area Under the Curves question

(i) Intersection of $y=4x^2$ and $y=4$

(ii) Intersection of $y=\frac{x^2}{9}$ and $y=4$

First integral

Second integral

More from Area Under the Curves

Area Under the Curves question

Area Under the Curves question

(i) Intersection of $y=4x^2$ and $y=4$

(ii) Intersection of $y=\frac{x^2}{9}$ and $y=4$

First integral

Second integral

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(i) Intersection of y=4x2y=4x^2y=4x2 and y=4y=4y=4

(ii) Intersection of y=x29y=\frac{x^2}{9}y=9x2​ and y=4y=4y=4

First integral

Second integral

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(i) Intersection of y=4x2y=4x^2y=4x2 and y=4y=4y=4

(ii) Intersection of y=x29y=\frac{x^2}{9}y=9x2​ and y=4y=4y=4

First integral

Second integral

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(i) Intersection of $y=4x^2$ and $y=4$

(ii) Intersection of $y=\frac{x^2}{9}$ and $y=4$

(i) Intersection of $y=4x^2$ and $y=4$

(ii) Intersection of $y=\frac{x^2}{9}$ and $y=4$