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The figure shows the parabola y² = 4x and the straight line y = x - 3…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

The figure shows the parabola $y^2 = 4x$ and the straight line $y = x - 3$ . The line cuts the parabola at two points: $P$ , above the $x$ -axis, and $Q$ , below it. The arc of the parabola between $P$ and $Q$ and the chord $PQ$ enclose a region $R$ , shaded in the figure.

The parabola y^2 = 4x, opening to the right from the origin, and the straight line y = x - 3, with axes and a light grid. The line cuts the parabola at P, up and to the right, and at Q, below the x-axis. The region between the arc of the parabola and the chord PQ is shaded; its left edge is the parabola. No tangent line is drawn.

A region like this can be cut into thin vertical strips or into thin horizontal strips. Look at the left edge of $R$ before you choose. With one choice a single integral is enough; with the other, the strips change their shape part of the way across, and the area needs two integrals.

A second straight line, parallel to $PQ$ , touches the parabola at a point $T$ on the arc between $P$ and $Q$ . Long before integrals were invented, Archimedes compared the area of a region like $R$ with the area of the triangle $PQT$ .

What is the area of $R$ ?
What is the area of the part of $R$ that lies above the $x$ -axis?
What is the $y$ -coordinate of $T$ ?
What is the area of the triangle $PQT$ ?
What is the ratio of the area of $R$ to the area of the triangle $PQT$ ?

Give every area in square units, to two decimals.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

The corners. On the line $x = y + 3$ , so $y^2 = 4(y + 3)$ , that is $y^2 - 4y - 12 = (y - 6)(y + 2) = 0$ . So $P = (9, 6)$ and $Q = (1, -2)$ .

(a) For $-2 \le y \le 6$ a horizontal strip runs from $x = \dfrac{y^2}{4}$ to $x = y + 3$ . So the area is $\displaystyle\int_{-2}^{6} \left(y + 3 - \frac{y^2}{4}\right) dy = \left[\frac{y^2}{2} + 3y - \frac{y^3}{12}\right]_{-2}^{6} = 18 - \left(-\frac{10}{3}\right) = \frac{64}{3} \approx 21.33$ square units.

(b) The part above the $x$ -axis is the part with $0 \le y \le 6$ : $\left[\dfrac{y^2}{2} + 3y - \dfrac{y^3}{12}\right]_0^6 = 18 + 18 - 18 = 18.00$ square units. The part below has area $\frac{10}{3} \approx 3.33$ square units.

(c) Differentiating $y^2 = 4x$ gives $2y\,\dfrac{dy}{dx} = 4$ , so the slope is $\dfrac2y$ . It equals the slope $1$ of $PQ$ when $y = 2$ . So $T = (1, 2)$ .

(d) $Q(1, -2)$ and $T(1, 2)$ lie on the vertical line $x = 1$ , so take $QT$ , of length $4$ , as the base. The height is the distance of $P(9, 6)$ from that line, $9 - 1 = 8$ . The area is $\dfrac12 \times 4 \times 8 = 16$ square units.

(e) $\dfrac{64/3}{16} = \dfrac43 \approx 1.33$ . This is Archimedes' result: a parabolic segment is always $\dfrac43$ of its triangle.

Check by vertical strips. For $0 \le x \le 1$ the strip runs from $-2\sqrt x$ to $2\sqrt x$ , and for $1 \le x \le 9$ from $x - 3$ to $2\sqrt x$ : $\displaystyle\int_0^1 4\sqrt x\,dx + \int_1^9 \left(2\sqrt x - x + 3\right) dx = \frac83 + \frac{56}{3} = \frac{64}{3}$ , the same area.

Watch out. With vertical strips it is easy to use $x - 3$ as the lower edge all the way from $x = 1$ to $x = 9$ and stop there. That leaves out the thin piece with $0 < x < 1$ , and gives only $\frac{56}{3} \approx 18.67$ square units.

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