Sample 2024-25 Class 12 Maths Q18 Solution: Area bounded by x² = 4y and the line y = 4

The question

A student observes an open-air Honeybee nest on the branch of a tree, whose plane figure is parabolic shape given by $x^2 = 4y$ ⁠. Then the area (in sq units) of the region bounded by parabola $x^2 = 4y$ and the line $y = 4$ is

(A) $\frac{32}{3}$

(B) $\frac{64}{3}$

(C) $\frac{128}{3}$

(D) $\frac{256}{3}$

Exam answer

Putting $y = 4$ in $x^2 = 4y$ ⁠: $x^2 = 16$ ⁠, so $x = \pm 4$ ⁠. The line meets the parabola at $(-4, 4)$ and $(4, 4)$ ⁠.

The region is symmetric about the $y$ -axis. For $x \ge 0$ ⁠, $x = 2\sqrt y$ ⁠, $0 \le y \le 4$ ⁠.

Why?

$x^2 = 4y$ is an upward parabola with vertex at the origin. Replacing $x$ by $-x$ leaves the equation unchanged, so the two halves are mirror images in the $y$ -axis.

For the right half take the positive root: $x = \sqrt{4y} = 2\sqrt y$ ⁠. The region runs from the vertex, $y = 0$ ⁠, up to the line $y = 4$ ⁠.

\begin{aligned} \text{Area} &= 2\int_0^4 2\sqrt y\,dy \\ &= 4\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^4 \\ &= \frac{8}{3}\left[y^{\frac{3}{2}}\right]_0^4 \\ &= \frac{8}{3}(8 - 0) = \frac{64}{3} \end{aligned}

Why?

With horizontal strips, the area of the right half is $\int_0^4 x\,dy$ ⁠; the factor 2 doubles it for the left half. Then the power rule:

\int\sqrt y\,dy = \frac{y^{\frac32}}{\frac32} = \frac{2}{3}y^{\frac32}

So $4 \times \frac23 = \frac83$ ⁠, and $4^{\frac32} = (\sqrt 4)^3 = 8$ ⁠.

\begin{aligned} 2\int_0^4 2\sqrt y\,dy &= 4\int_0^4 \sqrt y\,dy \\[4pt] 4 \times \frac23 &= \frac{8}{3} \\[4pt] \left[y^{\frac32}\right]_0^4 &= 4^{\frac32} - 0^{\frac32} \\ &= (\sqrt 4)^3 - 0 \\ &= 2^3 = 8 \\[4pt] \frac83 \times 8 &= \frac{64}{3} \end{aligned}

Watch out: forgetting the factor 2 for the left half gives $\frac{32}{3}$ ⁠, option (A).

Answer: (B) $\dfrac{64}{3}$ sq units

Where this comes from

Chapter 8, Application of Integrals: Area between a parabola and a line.

Question type: A region against the $y$ -axis

How to spot it: click to see

Type 2. A region against the $y$ -axis: horizontal strips

Spot it: the region is bounded by the $y$ -axis, or the limits are $y$ values, like the line $y = 3$ ⁠.

Here: The line is $y = 4$ ⁠, a $y$ limit. Write the parabola as $x = 2\sqrt y$ ⁠, integrate from $y = 0$ to 4, and double for the left half: $2\int_0^4 2\sqrt y\,dy = \frac{64}{3}$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 7, Integrals.

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Q18: Area bounded by x2=4yx^2 = 4yx2=4y and the line y=4y = 4y=4

Q18: Area bounded by $x^2 = 4y$ and the line $y = 4$