1. Calculate the area of the segment cut off from the parabola \(y = x^{2}/4a\) by the ordinate \(x = \xi\), and the length of the arc which bounds it.

2. Answer the same questions for the curve \(ay^{2} = x^{3}\), showing that the length of the arc is \[\frac{8a}{27} \left\{\left(1 + \frac{9\xi}{4a}\right)^{3/2} – 1\right\}.\]

3. Calculate the areas and lengths of the circles \(x^{2} + y^{2} = a^{2}\), \(x^{2} + y^{2} = 2ax\) by means of the formulae of §§ 145–146.

4. Show that the area of the ellipse \((x^{2}/a^{2}) + (y^{2}/b^{2}) = 1\) is \(\pi ab\).

5. Find the area bounded by the curve \(y = \sin x\) and the segment of the axis of \(x\) from \(x = 0\) to \(x = 2\pi\). [Here \(\Phi(x) = -\cos x\), and the difference between the values of \(-\cos x\) for \(x = 0\) and \(x = 2\pi\) is zero. The explanation of this is of course that between \(x = \pi\) and \(x = 2\pi\) the curve lies below the axis of \(x\), and so the corresponding part of the area is counted negative in applying the method. The area from \(x = 0\) to \(x = \pi\) is \(-\cos \pi + \cos 0 = 2\); and the whole area required, when every part is counted positive, is twice this, *i.e.* is \(4\).]

6. Suppose that the coordinates of any point on a curve are expressed as functions of a parameter \(t\) by equations of the type \(x = \phi(t)\), \(y = \psi(t)\), \(\phi\) and \(\psi\) being functions of \(t\) with continuous derivatives. Prove that if \(x\) steadily increases as \(t\) varies from \(t_{0}\) to \(t_{1}\), then the area of the region bounded by the corresponding portion of the curve, the axis of \(x\), and the two ordinates corresponding to \(t_{0}\) and \(t_{1}\), is, apart from sign, \(A(t_{1}) – A(t_{0})\), where \[A(t) = \int \psi(t)\phi'(t)\, dt = \int y \frac{dx}{dt}\, dt.\]

7. Suppose that \(C\) is a closed curve formed of a single loop and not met by any parallel to either axis in more than two points. And suppose that the coordinates of any point \(P\) on the curve can be expressed as in Ex. 6 in terms of \(t\), and that, as \(t\) varies from \(t_{0}\) to \(t_{1}\), \(P\) moves in the same direction round the curve and returns after a single circuit to its original position. Show that the area of the loop is equal to the difference of the initial and final values of any one of the integrals \[-\int y \frac{dx}{dt}\, dt,\quad \int x \frac{dy}{dt}\, dt,\quad \tfrac{1}{2} \int \left(x \frac{dy}{dt} – y \frac{dx}{dt}\right) dt,\] this difference being of course taken positively.

8. Apply the result of Ex. 7 to determine the areas of the curves given by \[(i) \frac{x}{a} = \frac{1 – t^{2}}{1 + t^{2}},\quad \frac{y}{a} = \frac{2t}{1 + t^{2}},\qquad (ii) x = a\cos^{3} t,\quad y = b\sin^{3} t.\]

9. Find the area of the loop of the curve \(x^{3} + y^{3} = 3axy\). [Putting \(y = tx\) we obtain \(x = 3at/(1 + t^{3})\), \(y = 3at^{2}/(1 + t^{3})\). As \(t\) varies from \(0\) towards \(\infty\) the loop is described once. Also \[\tfrac{1}{2} \int \left(y \frac{dx}{dt} – x \frac{dy}{dt}\right)\, dt = -\tfrac{1}{2} \int x^{2} \frac{d}{dt}\left(\frac{y}{x}\right)\, dt = -\tfrac{1}{2} \int \frac{9a^{2}t^{2}}{(1 + t^{3})^{2}}\, dt = \frac{3a^{2}}{2(1 + t^{3})},\] which tends to \(0\) as \(t \to \infty\). Thus the area of the loop is \(\frac{3}{2}a^{2}\).]

10. Find the area of the loop of the curve \(x^{5} + y^{5} = 5ax^{2}y^{2}\).

11. Prove that the area of a loop of the curve \(x = a\sin 2t\), \(y = a\sin t\) is \(\frac{4}{3}a^{2}\).

12. The arc of the ellipse given by \(x = a\cos t\), \(y = b\sin t\), between the points \(t = t_{1}\) and \(t = t_{2}\), is \(F(t_{2}) – F(t_{1})\), where \[F(t) = a\int \sqrt{1 – e^{2}\sin^{2} t}\, dt,\] \(e\) being the eccentricity. [This integral cannot however be evaluated in terms of such functions as are at present at our disposal.]

13. **Polar coordinates.** Show that the area bounded by the curve \(r = f(\theta)\), where \(f(\theta)\) is a one-valued function of \(\theta\), and the radii \(\theta = \theta_{1}\), \(\theta = \theta_{2}\), is \(F(\theta_{2}) – F(\theta_{1})\), where \( F(\theta) = \tfrac{1}{2} \int r^{2}\, d\theta\). And the length of the corresponding arc of the curve is \(\Phi(\theta_{2}) – \Phi(\theta_{1})\), where \[\Phi(\theta) = \int \sqrt{r^{2} + \biggl(\frac{dr}{d\theta}\biggr)^{2}}\, d\theta.\]

Hence determine (i) the area and perimeter of the circle \(r = 2a\sin\theta\); (ii) the area between the parabola \(r = \frac{1}{2}l\sec^{2} \frac{1}{2}\theta\) and its latus rectum, and the length of the corresponding arc of the parabola; (iii) the area of the limaçon \(r = a + b\cos\theta\), distinguishing the cases in which \(a > b\), \(a = b\), and \(a < b\); and (iv) the areas of the ellipses \(1/r^{2} = a\cos^{2} \theta + 2h\cos\theta\sin\theta + b\sin^{2} \theta\) and \(l/r = 1 + e\cos\theta\). [In the last case we are led to the integral \(\int \frac{d\theta}{(1 + e\cos\theta)^{2}}\), which may be calculated (cf. Ex. LIII. 4) by the help of the substitution \[(1 + e\cos\theta) (1 – e\cos\phi) = 1 – e^{2}.]\]

14. Trace the curve \(2\theta = (a/r) + (r/a)\), and show that the area bounded by the radius vector \(\theta = \beta\), and the two branches which touch at the point \(r = a\), \(\theta = 1\), is \(\frac{2}{3} a^{2}(\beta^{2} – 1)^{3/2}\).

15. A curve is given by an equation \(p = f(r)\), \(r\) being the radius vector and \(p\) the perpendicular from the origin on to the tangent. Show that the calculation of the area of the region bounded by an arc of the curve and two radii vectores depends upon that of the integral \(\frac{1}{2} \int \frac{pr\, dr}{\sqrt{r^{2} – p^{2}}}\).