Further Pure | Topic 5: Differentiation and Integration

Curriculum Objectives:

• obtain an expression for $\frac{{{d^2}y}}{{d{x^2}}}$ in cases where the relation between y and x is defined parametrically or implicitly
• derive and use reduction formulae for the evaluation of definite integral in simple cases
• use of integration to find: mean values and centroids of two or three dimensional figures ( where equations are expressed in Cartesian co-ordinates, including the use of a parameter) using strips, discs and shells as appropriate, arc lengths ( for curves in Cartesian co-ordinates or in polar form)
• surface area of revolution about one of the axes ( for curves in Cartesian co-ordinates but not for those in polar form )

The second derivative of an implicitly defined function

You are reminded that an implicit function is given by:

$f({x_1},{x_2},...,{x_n}) = 0$

However we are concerned only with the case where there are two variables. For example, the function defining a circle, with the origin as the centre and c as the radius:

${x^2} + {y^2} = {c^2}$

Using this relatively simple example let me illustrate the method of arriving at the second derivative with respect to either variable. We are going to use x, x', x'', y, y', y'', as opposed to $x,\frac{{dx}}{{dy}},\frac{{{d^2}x}}{{d{y^2}}},y,\frac{{dy}}{{dx}},\frac{{{d^2}y}}{{d{x^2}}}$, on account of the, being easier to work with in this case, as you will see.

Differentiating once with respect to x, as per the implicit differentiation you must have learnt in your A Levels P3.

$2x + 2yy' = 0$
$y' =  - \frac{x}{y}$

Differentiating again with respect to x, we find that the term 2x is easily differentiated; differentiation of yy' however is notion as yet foreign. For this we recognise that y' must too be a function of x and that the product rule is then applicable here, and that:

$(y')' = y''$ or $\frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{{d^2}y}}{{d{x^2}}}$

The differentiation of yy' is then:

$(yy')' = yy'' + {(y')^2}$

Then the second differentiation of the curve should yield:

$2 + 2yy'' + 2{(y')^2} = 0$
$1 + yy'' + {(y')^2} = 0$
$yy'' =  - 1 - {(y')^2}$
$yy'' =  - 1 - {\left( { - \frac{x}{y}} \right)^2}$
$y'' =  - \frac{1}{y} - \frac{{{x^2}}}{{{y^3}}}$
$y'' = \frac{{ - ({y^2} + {x^2})}}{{{y^3}}}$

A Further substitution could yield:

$y'' =  - \frac{{{c^2}}}{{{y^3}}}$


Ex. (October/November 2011, Paper 12, Q 5)

The point P (2, 1) lies on the curve with the equation:

${x^3} - 2{y^3} = 3xy$

Find
(I) the value of $\frac{{dy}}{{dx}}$ at P
(II) the value of $\frac{{{d^2}y}}{{d{x^2}}}$ at P

Sol.

(I) We differentiate the equation, to get:

$3{x^2} - 6{y^2}y' = 3(y + xy')$

Putting in the co-ordinates of P

$3{(2)^2} - 6{(1)^2}y' = 3(1 + 2y')$
$12 - 6y' = 3 + 6y'$
$y' = \frac{3}{4}$

(II) Differentiating we  get

$6x - 6({y^2}y'' + 2y{(y')^2}) = 3y' + 3(y' + xy'')$
$6(2) - 6\left( {y'' + \frac{9}{8}} \right) = \frac{9}{4} + \frac{9}{4} + 6y''$
$12y'' = \frac{3}{4}$
$y'' = \frac{1}{{16}}$

The Second Derivative of a parametrically defined function

Here, too, the method will be explained with the aid of examples, although no particular notation is given precedence.

Ex. (Oct/Nov 2013 Paper 13, Q4)

A curve has the parametric equations:

$x = 2\theta  - \sin (2\theta )$  $y = 1 - \cos (2\theta )$ for  $- 3\pi  \le \theta  \le 3\pi$

Show that

$\frac{{dy}}{{dx}} = \cot \theta $

expect for certain values of $\theta $, which should be stated.

Sol.

By differentiating with respect to $\theta $ we have that:

$\frac{{dx}}{{d\theta }} = 2 - 2\cos (2\theta )$ and $\frac{{dy}}{{d\theta }} = 2\sin (2\theta )$

We further realise that:

$\frac{{dy}}{{dx}} = \frac{{2\sin (2\theta )}}{{2 - 2\cos (2\theta )}}$

$ = \frac{{\sin (2\theta )}}{{1 - \cos (2\theta )}}$

$ = \frac{{2\sin \theta cos\theta }}{{2{{\sin }^2}\theta }}$

$ = \frac{{\cos \theta }}{{\sin \theta }}$

$ = \cot \theta $

For all the integral multiples of pi will leave us with a zero denominator, n * pi, where n is an integer, are the values of $\theta $ for which this derivative is undefined

Find the value of $\frac{{{d^2}y}}{{d{x^2}}}$ for $\theta $ = (pi)/4

For this we must make the following considerations:

$\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{dx}}\left( {\cot \theta } \right) = \frac{d}{{d\theta }}(\cot \theta ) \cdot \frac{{d\theta }}{{dx}}$

Recall that the differentiation of $\cot \theta $ is $ - {\csc ^2}\theta $ a that

$\frac{{d\theta }}{{dx}} = \frac{1}{{\frac{{dx}}{{d\theta }}}}$

$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - {{\csc }^2}\theta }}{{2 - 2\cos 2\theta }}$

$ = \frac{{ - {{\csc }^2}\theta }}{{4{{\sin }^2}\theta }}$

$ =  - \frac{1}{{4{{\sin }^4}\theta }}$

Putting in the value of $\theta $, we find the value -1

Reduction Formula

In P3, oft times, question involved integrating by parts twice to achieve the desired result. Consider the following example:

$\int {{x^2}{e^x}dx} $

And we let:

$u = {x^2}$ $v' = {e^x}$

$u' = 2x$ $v = {e^x}$

Then we have:

$\int {u\frac{{dv}}{{dx}}dx}  = uv - \int {v\frac{{du}}{{dx}}dx} $

$\int {{x^2}{e^x}dx}  = {x^2}{e^x} - 2\int {{e^x}x} dx$

Now if I were to let $I_2$ and $I_1$ be:

${I_2} = \int {{x^2}{e^x}} dx$

${I_1} = \int {{x}{e^x}} dx$

Which are thusly related:

${I_2} = {x^2}{e^x} - 2{I_1}$

If instead we begin with the general term n.

${I_n} = \int {{x^n}{e^x}dx} $

We have:

$u = {x^n}$  $v' = {e^x}$

$u' = n{x^{n - 1}}$ $v = {e^x}$

$\int {{x^n}{e^x}dx = {x^n}{e^x} - n\int {{x^{n - 1}}{e^x}dx} } $

${I_n} = {x^n}{e^x} - n{I_{n - 1}}$

This is a reduction formula, which in essence reduces a rather complicated integral to a simpler one.

Say, now that we wanted to find $I_3$. We have:

${I_3} = {x^3}{e^x} - 3{I_2} = {x^3}{e^x} - 3({x^2}{e^x} - 2{I_1}) = {x^3}{e^x} - 3({x^2}{e^x} - 2(x{e^x} - {I_0}))$

${I_0} = \int {{x^0}{e^x}dx}  = \int {{e^x}dx}  = {e^x}$

Then you input the value of $I_0$ to get:

${I_3} = {x^3}{e^x} - 3{x^2}{e^x} + 6x{e^x} - 6{e^x} + k$

It is customary not to introduce the constant of integration, until the final integration has been completed.

This topic will require relentless practice to master, as is true for most of calculus.

While the integrating by parts methods has been used to introduce the topic and you will find it insinuated in almost all the question on the topics, other methods can be used to derive reduction formulae. Examples in this regard could be of differentiation and identities.

Ex.

Find the reduction formula for:

${I_n} = \int_0^{\frac{\pi }{4}} {{{\tan }^n}x} dx$

We rewrite and then use a well known identity to establish the result.

 ${I_n} = \int_0^{\frac{\pi }{4}} {{{\tan }^n}xdx} $
${I_n} = \int_0^{\frac{\pi }{4}} {{{\tan }^{n - 2}}x \cdot {{\tan }^2}x} dx$
${I_n} = \int_0^{\frac{\pi }{4}} {{{\tan }^{n - 2}}x \cdot } ({\sec ^2}x - 1)dx$
${I_n} = \int_0^{\frac{\pi }{4}} {{{\tan }^{n - 2}}x} {\sec ^2}xdx - \int_0^{\frac{\pi }{4}} {{{\tan }^{n - 2}}x} dx$
${I_n} = \left[ {\frac{{{{\tan }^{n - 1}}x}}{{n - 1}}} \right]_0^{\frac{\pi }{4}} - {I_{n - 2}}$
${I_n} = \frac{1}{{n - 1}} - {I_{n - 2}}$

Provided of course that $n \ge 2$

Let us attempt a couple of past paper questions before moving on.

Ex. (October/November 2012 Paper 13 Question 11)

Show that ${\int {x\left( {1 - {x^2}} \right)} ^{\frac{1}{2}}}dx =  - \frac{1}{3}{\left( {1 - {x^2}} \right)^{\frac{3}{2}}} + c$

Given that ${I_n} = \int_0^1 {{x^n}{{(1 - {x^2})}^{\frac{1}{2}}}dx} $ prove that for $n \ge 2$

$\left( {n + 2} \right){I_n} = \left( {n - 1} \right){I_{n - 2}}$

Use the substitution x = sin u to show that:

${\int_0^1 {\left( {1 - {x^2}} \right)} ^{\frac{1}{2}}}dx = \frac{ \pi }{4}$

Find ${I_4}$
Sol.

For the first part we let:

$u = 1 - {x^2}$
$\frac{{du}}{{dx}} =  - 2x$

We rewrite the given integral thusly:

$ - \frac{1}{2}\int { - 2x{{(1 - {x^2})}^{\frac{1}{2}}}dx} $

Then substituting u and u' into the integral

$ - \frac{1}{2}\int {\frac{{du}}{{dx}}{{(u)}^{\frac{1}{2}}}dx}  =  - \frac{1}{2}\int {{u^{\frac{1}{2}}}} du =  - \frac{1}{2}\left[ {\frac{2}{3}{u^{\frac{3}{2}}}} \right] =  - \frac{1}{3}{\left( {1 - {x^2}} \right)^{\frac{3}{2}}} + c$

Obviously, in having us evaluate a rather simple integral, the examiner wishes us to use it in the next part, but disregarding that at this point, if you consider the following:

$u = {x^n}$      $v' = {\left( {1 - {x^2}} \right)^{\frac{1}{2}}}$

The differentiation of u is easy enough, the integration of v' however is slightly difficult and will not help prove the relationship that is required. Yet the integral is evaluated so to impress upon you the fact, that seldom, does the most apparent method is correct method in a further mathematics paper.

We begin with the substitution:

$x = \sin u$
$\frac{{dx}}{{du}} = \cos u$
$dx = \cos u  du$

Also we have:

$u = {\sin ^{ - 1}}x$

By making the substitutions and using the identity linking the squares of sine and cosine, and unity, we find that:

$\sqrt {1 - {{\sin }^2}u}  = \sqrt {{{\cos }^2}x}  = \cos x$ 

The integral becomes, therefore:

$\int {\sqrt {1 - {x^2}} dx}  = \int {\cos u} dx = \int {\cos u \cdot \cos udu = \int {{{\cos }^2}udu} } $

We know that:

${\cos ^2}u = \frac{{\cos 2u}}{2} + \frac{1}{2}$

Therefore the integral becomes

$\int {\left( {\frac{{\cos 2u}}{2} + \frac{1}{2}} \right)} du$

Integrating it using the rules you must have studied in P3, we get:

$\int {\left( {\frac{{\cos 2u}}{2} + \frac{1}{2}} \right)} du = \frac{{\sin (2u)}}{4} + \frac{u}{2}$

The constant of integration has been disregarded at this point. x is now substituted back. We get:

$\frac{1}{4}\sin (2u) + \frac{1}{2}u = \frac{1}{4}\sin (2{\sin ^{ - 1}}x) + \frac{1}{2}{\sin ^{ - 1}}x$

Using the following identity we may clean the integral up somewhat.

$\sin (2{\sin ^{ - 1}}x) = 2x\sqrt {1 - {x^2}} $

Finally we write that:

$\int {\sqrt {1 - {x^2}} } dx = \frac{x}{2}\sqrt {1 - {x^2}} + \frac{1}{2}{\sin ^{ - 1}}x + c$

This evidently doesn't help us in the least. We know attempt the second part using the integral we have proven in the first part of the question. We must adjust the $I_n$ so that the consequent integration by parts involves:

$\int {x\sqrt {1 - {x^2}} } dx$

Hence we manipulate the given integral thusly:

$\int_0^1 {{x^{n - 1}} \cdot x\sqrt {1 - {x^2}} dx} $

And let:

$u = {x^{n - 1}}$    $v' = x\sqrt {1 - {x^2}} $
$u' = \left( {n - 1} \right){x^{n - 2}}$   $v =  - \frac{1}{3}{\left( {1 - {x^2}} \right)^{\frac{3}{2}}} + c$

The formula for integration by parts then states:

${I_n} = \left[ {\left( {{x^{n - 1}}} \right)\left( { - \frac{1}{3}{{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}} \right)} \right]_0^1 - \int_0^1 {(n - 1){x^{n - 2}} \cdot \left( { - \frac{1}{3}{{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}} \right)} dx$

${I_n} = \left[ {\left( {{x^{n - 1}}} \right)\left( { - \frac{1}{3}{{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}} \right)} \right]_0^1 + \frac{1}{3}(n - 1)\int_0^1 {{x^{n - 2}} \cdot {{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}} dx$

${I_n} = \frac{1}{3}(n - 1)\int_0^1 {{x^{n - 2}} \cdot {{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}} dx$

$3{I_n} = (n - 1)\int_0^1 {{x^{n - 2}} \cdot {{\left( {1 - {x^2}} \right)}^{1 + \frac{1}{2}}}} dx$

$3{I_n} = (n - 1)\int_0^1 {{x^{n - 2}} \cdot \left( {1 - {x^2}} \right){{\left( {1 - {x^2}} \right)}^{\frac{1}{2}}}} dx$

$3{I_n} = (n - 1)\int_0^1 {{x^{n - 2}} \cdot {{\left( {1 - {x^2}} \right)}^{\frac{1}{2}}}} dx - (n - 1)\int_0^1 {{x^n} \cdot {{\left( {1 - {x^2}} \right)}^{\frac{1}{2}}}} dx$

$3{I_n} = (n - 1)\int_0^1 {{x^{n - 2}} \cdot {{\left( {1 - {x^2}} \right)}^{\frac{1}{2}}}} dx - (n - 1){I_n}$

$(n + 2){I_n} = (n - 1)\int_0^1 {{x^{n - 2}} \cdot {{\left( {1 - {x^2}} \right)}^{\frac{1}{2}}}} dx = (n - 1){I_{n - 2}}$

The indefinite integral has already been evaluated, you need only complete the calculation with the bounds given, and it's easily proven.

$6{I_4} = 3{I_2} \Rightarrow {I_4} = \frac{1}{2}{I_2}$

$4{I_2} = {I_0} \Rightarrow {I_2} = \frac{1}{4}{I_0}$

${I_4} = \frac{1}{2}\left( {\frac{1}{4}{I_0}} \right) = \frac{1}{8}{I_0} = \frac{\pi }{{32}}$

Ex. (October/November 2009; Paper 01; Q6)

Show that

$\frac{d}{{dx}}\left[ {{x^{n - 1}}\sqrt {4 - {x^2}} } \right] = \frac{{4(n - 1){x^{n - 2}}}}{{\sqrt {4 - {x^2}} }} - \frac{{n{x^n}}}{{\sqrt {4 - {x^2}} }}$

Let

${I_n} = \int_0^1 {\frac{{{x^n}}}{{\sqrt {4 - {x^2}} }}} dx$

where $n \ge 0$. Prove that

$n{I_n} = 4(n - 1){I_{n - 2}} - \sqrt 3 $

for $n \ge 2$

Given that ${I_0} = \frac{1}{6}\pi $, find ${I_4}$, leaving your answer in exact form.

Sol.

$\frac{d}{{dx}}\left[ {{x^{n - 1}}\sqrt {4 - {x^2}} } \right] = {x^{n - 1}}\frac{d}{{dx}}\left[ {\sqrt {4 - {x^2}} } \right] + \sqrt {4 - {x^2}} \frac{d}{{dx}}\left[ {{x^{n - 1}}} \right]$

$ = {x^{n - 1}}\left[ {\frac{1}{2} \cdot \frac{1}{{\sqrt {4 - {x^2}} }} \cdot \left( { - 2x} \right)} \right] + \sqrt {4 - {x^2}} \left[ {\left( {n - 1} \right){x^{n - 2}}} \right]$

$ =  - \frac{{{x^n}}}{{\sqrt {4 - {x^2}} }} + \sqrt {4 - {x^2}} \left[ {\left( {n - 1} \right){x^{n - 2}}} \right]$

$ =  - \frac{{{x^n}\sqrt {4 - {x^2}} }}{{4 - {x^2}}} + \sqrt {4 - {x^2}} (n - 1){x^{n - 2}}$

$ = \sqrt {4 - {x^2}} \left[ { - \frac{{{x^n}}}{{4 - {x^2}}} + (n - 1){x^{n - 2}}} \right]$

$ = \sqrt {4 - {x^2}} \left[ {\frac{{ - {x^n} + (4 - {x^2})(n - 1){x^{n - 2}}}}{{4 - {x^2}}}} \right]$

$ = \sqrt {4 - {x^2}} \left[ {\frac{{ - {x^n} + 4(n - 1){x^{n - 2}} - n{x^{n - 2 + 2}} + {x^{n - 2}}}}{{4 - {x^2}}}} \right]$

$ = \sqrt {4 - {x^2}} \left[ {\frac{{4(n - 1){x^{n - 2}} - n{x^n}}}{{4 - {x^2}}}} \right]$

Furthermore, you need only cancel out the term $\sqrt {4 - {x^2}} $ to achieve the required result.

For the next part, we integrate the relationship we have already proven.

$\int_0^1 {\frac{d}{{dx}}\left[ {{x^{n - 1}}\sqrt {4 - {x^2}} } \right]} dx = \int_0^1 {\frac{{4(n - 1){x^{n - 2}}}}{{\sqrt {4 - {x^2}} }}} dx - \int_0^1 {\frac{{n{x^n}}}{{\sqrt {4 - {x^2}} }}} dx$

$\left[ {{x^{n - 1}}\sqrt {4 - {x^2}} } \right]_0^1 = 4(n - 1)\int_0^1 {\frac{{{x^{n - 2}}}}{{\sqrt {4 - {x^2}} }}dx - n\int_0^1 {\frac{{{x^n}}}{{\sqrt {4 - {x^2}} }}} } dx$

$\sqrt 3  = 4(n - 1){I_{n - 2}} - n{I_n}$

$n{I_n} = 4(n - 1){I_{n - 2}} - \sqrt 3 $

Centroid

The centroid of an object is such a point, that it is the average ( or arithmetic mean ) of all the points in the object. Say an object consists of n number of points, and $x_i$ is the distance of the point from some reference point in the x-axis, then the distance of the x coordinate of the centroid of the object from that reference point is given by:

${C_x} = \frac{{{x_1} + {x_2} + {x_3}...{x_n}}}{n}$

This can obviously be extended to any number of finite dimensions. If you have completed the M2 unit of the CIE A levels Mathematics Course, you might surmise some similitude centroids have with centres of mass. Assuming uniform density, an object has the same point for its centroid and centre of mass. You will not be questioned about this in the pure paper. For the applied paper we will study this topic in greater detail.

The formulae relevant to the pure paper, are as follows.

For a curve y = f(x) defined between a <  x <  b, the x and y co-ordinates of the centroid is given by:

${C_x} = \frac{{\int_a^b {xy} {\kern 1pt} dx}}{{\int_a^b y {\kern 1pt} dx}}$ ${C_y} = \frac{1}{2}\frac{{\int_a^b {{y^2}} dx}}{{\int_a^b y {\kern 1pt} dx}}$

Ex. Find the co-ordinates of the centroid of the curve $y = \sqrt {1 - x} $ for $0 \le x \le 1$ 

Sol.

We note:

$\int_0^1 {ydx}  = \int_0^1 {{{(1 - x)}^{\frac{1}{2}}}} dx = \left[ { - \frac{2}{3}{{(1 - x)}^{\frac{3}{2}}}} \right]_0^1 = \frac{2}{3}$

We also note that:

$\int_0^1 {xydx}  = \int_0^1 {x{{(1 - x)}^{\frac{1}{2}}}dx}  =  - \int_0^1 {(1 - u){{(u)}^{\frac{1}{2}}}du} $

u = 1 - x

$\int_0^1 {\left( {{u^{\frac{3}{2}}} - {u^{\frac{1}{2}}}} \right)} du = \frac{4}{{15}}$

${C_x} = \frac{{\frac{4}{{15}}}}{{\frac{2}{3}}} = \frac{2}{5}$

$\int_0^1 {{y^2}} dx = \int_0^1 {\left( {1 - x} \right)dx}  = \frac{1}{2}$

${C_y} = \frac{1}{2} \cdot \frac{{\frac{1}{2}}}{{\frac{2}{3}}} = \frac{3}{8}$

Mean Value of a function

For a function y = f(x), defined on the range a < x < b, the mean value of the function is given by:

$\bar y = \frac{1}{{b - a}}\int_0^1 {y{\kern 1pt} dx} $

Ex. (October/November 2012, Paper 12, Question Number 2)

The curve has, for $0 \le x \le 4$, the equation

$y = 2{x^{{\textstyle{1 \over 2}}}}$

Find

(I) the Mean value of y with respect to x
(II) the y-co-ordinate of the centroid of the region enclosed by the curve, the line x = 4 the x-axis

Sol.

(I) the mean value of function is given by:

$\bar y = \frac{1}{{b - a}}\int_0^4 y \,dx = \frac{1}{4}\int_0^4 {2{x^{\frac{1}{2}}}} {\kern 1pt} dx = \frac{1}{2}\left[ {\frac{2}{3}{x^{\frac{3}{2}}}} \right]_0^4 = \frac{8}{3}$

(II) for the y co-ordinate of the centroid, we note:

$\int_a^b {y{\kern 1pt} dx}  = \int_0^4 {2{x^{\frac{1}{2}}}} {\kern 1pt} dx = \left[ {\frac{4}{3}{x^{\frac{3}{2}}}} \right]_0^4 = \frac{{32}}{3}$

$\int_a^b {{y^2}dx}  = \int_0^4 {4x} {\kern 1pt} dx = 2\left[ {{x^2}} \right]_0^4 = 32$

The y co-ordinate is then simply:

$ = \frac{1}{2}\left[ {\frac{{32 \times 3}}{{32}}} \right] = \frac{3}{2}$

Arc Length

The arc length of a curve defined in rectangular, parametric, or polar equations, is found by the application of the following formulae (provided the function is differentiable over the given range):

1) For the function y = f(x), for the region a < x < b

$S = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx$

2) For the function x = f(y) (which is the inverse of the above function), defined for c < y < d

$S = \int_c^d {\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} } dy$

3) For the function defined parametrically, y = f(t); x = f(t), for the region e < t < f

$S = \int_e^f {\sqrt {{{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dx}}{{dt}}} \right)}^2}} } dt$

4) For the polar function defined r = f($ \theta $), for g < $ \theta $ < h,

$S = \int_g^h {\sqrt {{{\left( r \right)}^2} + {{\left( {\frac{{dr}}{{d\theta }}} \right)}^2}} d\theta } $

Area of Surface of Revolution

For the function $y = f(x)$, defined for a < x < b, is rotated through $2\pi$, about the x-axis,the area of the surface formed is given by:

$Surface\,Area = \int_a^b {2\pi y} \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \,dx$

For the function, defined for c < y < d rotated through $2\pi$, about the y-axis, the area of surface generated is given by:

$Surface\,Area = \int_c^d {2\pi x} \sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} \,dy$

For the function defined parametrically ( y = f(t); x = g(t) ), defined for e < t < f, the area of surface generated is given by:

$Surface\,Area = \int_c^d {2\pi y} \sqrt {{{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dx}}{{dt}}} \right)}^2}} \,dt$

By replacing y with x in the above formula, you get the formula for the parametric analogue of the function generated by rotating about x-axis.

Ex. (October/November 2012, Paper 12, Question 8)

The curve, C, has the parametric equations:

$x = {\textstyle{1 \over 3}}{t^3} - \ln t$
$y = {\textstyle{4 \over 3}}{t^{\frac{3}{2}}}$

for $1 \le t \le 3$. Find the arc length of C.
Find also the area of the surface generated when C is rotated through $2\pi# radians about the x-axis.

Sol.

We find the derivatives, with respect to t.

$\dot x = {t^2} - \frac{1}{t} = \frac{{{t^3} - 1}}{t}$
$\dot y = 2{t^{\frac{1}{2}}}$

We find the squares:

${\left( {\dot x} \right)^2} = {\left( {\frac{{{t^3} - 1}}{t}} \right)^2} = \frac{{{t^9} - 2{t^3} + 1}}{{{t^2}}}$

${\left( {\dot y} \right)^2} = 4t$

Performing the root extraction of the sum of the two squares of the derivatives:


$\sqrt {{{\left( {\dot x} \right)}^2} + {{\left( {\dot y} \right)}^2}}  = \sqrt {\frac{{{t^9} - 2{t^3} + 1}}{{{t^2}}} + 4t}  = \sqrt {\frac{{{t^9} + 2{t^3} + 1}}{{{t^2}}}} $

$ = \sqrt {\frac{{{{({t^3} + 1)}^2}}}{{{t^2}}}}  = \frac{{{t^3} + 1}}{t} = {t^2} + \frac{1}{t}$

Now we need only integrate this result, from 1 to 3.

For the area of the surface we have:

$\int_a^b {2\pi y\sqrt {{{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dx}}{{dt}}} \right)}^2}} {\kern 1pt} dt} $

We have already evaluated the following:

${\sqrt {{{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dx}}{{dt}}} \right)}^2}} {\kern 1pt} }$

We have therefore:

$2\pi \int_1^3 {y\left( {{t^2} + \frac{1}{t}} \right)\,dt} $

$2\pi \int_1^3 {\frac{4}{3}{t^{\frac{3}{2}}}\left( {{t^2} + \frac{1}{t}} \right)\,dt}  = \frac{{8\pi }}{3}\int_1^3 {{t^{\frac{7}{2}}}}  + {t^{\frac{1}{2}}}\,dt$

The integral is simple enough, and hence left to the reader.