Exercise 8.1 · Q4 (MCQ): Area between $y^2 = 4x$⁠, the $y$-axis and $y = 3$

Basics you need

Solution

1. Sketch and identify the region

   Putting $y = 3$ in $y^2 = 4x$⁠: $9 = 4x$⁠, so $x = \frac94$⁠. The line $y = 3$ meets the parabola at $P\left(\frac94, 3\right)$⁠, and it meets the $y$-axis at $(0, 3)$⁠. The parabola meets the $y$-axis ($x = 0$⁠) only at $O(0, 0)$⁠.

   So the three boundaries enclose the region with corners $O$⁠, $(0, 3)$ and $P$⁠: on the left the $y$-axis from 0 to 3, on top the line $y = 3$ from $x = 0$ to $x = \frac94$⁠, and on the right the arc of the parabola from $P$ back down to $O$⁠.

   Below the $x$-axis the parabola and the $y$-axis also lie side by side, but nothing closes that part off at the bottom, so it is not a bounded region. The only region the three given boundaries enclose is the one from $y = 0$ to $y = 3$⁠.
   Where a horizontal line $y = k$ meets a curve, put $y = k$ in its equation. Basics: reading the edges from the equation

2. Write the curve as $x$ in terms of $y$

   $$y^2 = 4x \;\Rightarrow\; x = \frac{y^2}{4}.$$
   A horizontal strip at height $y$ starts on the $y$-axis ($x = 0$⁠) and ends on the parabola, so its length is the parabola's $x$ at that height, $\frac{y^2}{4}$⁠, which is $\ge 0$⁠. One formula covers every strip from $y = 0$ to $y = 3$⁠.
   Turn your head sideways: $x$ is now the "height" of the strip. Basics: strips can be horizontal

3. Write the area as an integral

   $$\text{Area} = \int_0^3 x\,dy = \int_0^3\frac{y^2}{4}\,dy.$$
   Each strip has length $x$ and width $dy$⁠, so area $x\,dy$⁠; the strips run from $y = 0$ (the vertex $O$⁠) to $y = 3$ (the line).
   Section 8.2: $A = \int_c^d x\,dy$⁠. Lesson: horizontal strips

4. Integrate

   $$\int_0^3\frac{y^2}{4}\,dy = \Big[\frac{y^3}{12}\Big]_0^3 = \frac{27}{12} - 0 = \frac94.$$
   $\int y^2\,dy = \frac{y^3}{3}$⁠, and the factor $\frac14$ makes it $\frac{y^3}{12}$⁠. Then $3^3 = 27$ and $\frac{27}{12} = \frac{9}{4}$ after dividing top and bottom by 3.
   Formula (i), Section 7.2; Theorem 2, Section 7.8.3. Basics: a definite integral is a number

5. Check with vertical strips

   The region and the area under the upper half of the parabola, $y = 2\sqrt{x}$ from $x = 0$ to $x = \frac94$⁠, together fill the rectangle $0 \le x \le \frac94$⁠, $0 \le y \le 3$⁠, whose area is $\frac94 \times 3 = \frac{27}{4}$⁠. The area under the curve is

   $$\begin{aligned} \int_0^{9/4}2\sqrt{x}\,dx &= \Big[\frac43x^{3/2}\Big]_0^{9/4} \\ &= \frac43\cdot\frac{27}{8} = \frac92. \end{aligned}$$

   So the region has area $\frac{27}{4} - \frac92 = \frac{27}{4} - \frac{18}{4} = \frac94$⁠, the same. The correct option is (B).

   The arc $OP$ cuts the rectangle into two parts: the part left of it (our region) and the part under it. $\left(\frac94\right)^{3/2} = \left(\frac32\right)^3 = \frac{27}{8}$⁠. Two different methods agreeing confirms $\frac94$⁠.
   Section 8.2: $A = \int_a^b y\,dx$ for the part under the curve. Lesson: vertical strips

6. Examine every option

   A quick bound: the chord from $O$ to $P$⁠, the line $x = \frac{3y}{4}$⁠, lies to the right of the arc, since $\frac{3y}{4} \ge \frac{y^2}{4}$ for $0 \le y \le 3$⁠. So the region lies inside the triangle $O$⁠, $(0, 3)$⁠, $P$⁠, of area $\frac12\cdot\frac94\cdot 3 = \frac{27}{8} = 3.375$⁠.

   Option	Value	Verdict
   (A) $2$	2	Not the integral $\frac94$
   (B) $\frac{9}{4}$	2.25	Correct: $\int_0^3\frac{y^2}{4}\,dy$
   (C) $\frac{9}{3}$	3	Not the integral $\frac94$
   (D) $\frac{9}{2}$	4.5	Bigger than the triangle; it is the area under the curve

   (D) $\frac92$ is the trap: it is $\int_0^{9/4}2\sqrt{x}\,dx$⁠, the area between the parabola and the $x$-axis, the other part of the rectangle in Step 5. It is also larger than 3.375, so it cannot fit inside the triangle. (A) and (C) pass the size bound but are not the value of the integral; neither comes from a correct method.
   Read which axis the region lies against before choosing $dx$ or $dy$⁠. Lesson: why the right option is right

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