NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Further areas between two curves
Lesson

Following on from here, we will now explore some further integrations to find areas between curves. 

Example 1

We begin with the lens shape formed between two circles, $x^2+y^2=1$x2+y2=1 and $x^2+(y-1)^2=1$x2+(y1)2=1. The graphs are shown below.

Both circles have radius $1$1. One is centred at the point $(0,0)$(0,0) and the other at the point $(0,1)$(0,1). We will need the lower half of the top circle and the upper half of the lower circle in the form of functions. These are respectively, $y(x)=1-\sqrt{1-x^2}$y(x)=11x2 and $y(x)=\sqrt{1-x^2}$y(x)=1x2.

To find the enclosed area we calculate

$\int_a^b\ \sqrt{1-x^2}-\left(1-\sqrt{1-x^2}\right)\ \mathrm{d}x$ba 1x2(11x2) dx

where the terminals $a$a and $b$b have to be determined by finding the intersection points of the two semicircles.

At the intersection points, $\sqrt{1-x^2}=1-\sqrt{1-x^2}$1x2=11x2. Therefore, $2\sqrt{1-x^2}=1$21x2=1and then

$1-x^2$1x2 $=$= $\frac{1}{4}$14
$x^2$x2 $=$= $\frac{3}{4}$34
$x$x $=$= $\pm\frac{\sqrt{3}}{2}$±32

 

Because of the symmetry involved, we can simplify the calculation slightly by evaluating the integral from $0$0 to $\frac{\sqrt{3}}{2}$32. Thus, 

$\text{Area}=2\int_0^{\frac{\sqrt{3}}{2}}\ \sqrt{1-x^2}-\left(1-\sqrt{1-x^2}\right)\ \mathrm{d}x=2\int_0^{\frac{\sqrt{3}}{2}}\ 2\sqrt{1-x^2}-1\ \mathrm{d}x$Area=2320 1x2(11x2) dx=2320 21x21 dx

We make the trigonometric substitution $x=\sin\theta$x=sinθ. Then, $\frac{\mathrm{d}x}{\mathrm{d}\theta}=\cos\theta$dxdθ=cosθ and we can put $\cos\theta\ \mathrm{d}\theta$cosθ dθ instead of $\mathrm{d}x$dx. The terminals of the integral become $\theta=0$θ=0 and $\theta=\frac{\pi}{3}$θ=π3 (because $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$sinπ3=32). Thus, 

$\text{Area}$Area $=$= $2\int_0^{\frac{\pi}{3}}\ (2\cos\theta-1)\cos\theta\ \mathrm{d}\theta$2π30 (2cosθ1)cosθ dθ
  $=$=
$2\int_0^{\frac{\pi}{3}}\ 2\cos^2\theta-\cos\theta\ \mathrm{d}\theta$2π30 2cos2θcosθ dθ

Here, we make use of the identity $\cos2\theta\equiv2\cos^2\theta-1$cos2θ2cos2θ1 in the form $2\cos^2\theta\equiv\cos2\theta+1$2cos2θcos2θ+1. So,

$\text{Area}$Area $=$= $2\int_0^{\frac{\pi}{3}}\ \cos2\theta+1-\cos\theta\ \mathrm{d}\theta$2π30 cos2θ+1cosθ dθ
  $=$= $2\left[\frac{1}{2}\sin2\theta+\theta-\sin\theta\right]_0^{\frac{\pi}{3}}$2[12sin2θ+θsinθ]π30
  $=$= $2\left[\frac{\sqrt{3}}{4}+\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right]$2[34+π332]
  $=$= $\frac{2\pi}{3}-\frac{\sqrt{3}}{2}$2π332
  $\approx$ $1.228$1.228

 

One could devise similar problems with varying levels of difficulty by moving the centre of one of the circles or by changing the radius of one of them. The following example involves a unit circle and a horizontal line. Again, a lens-shaped area is formed between the graphs.

Example 2

We wish to find the area bounded by the semicircle $f(x)=\sqrt{1-x^2}$f(x)=1x2 and the line $g(x)=\frac{21}{29}$g(x)=2129. (We contrived this value of $g(x)$g(x) to make the intersection points with the semicircle rational.)

You should check that the intersection points are $\left(-\frac{20}{29},\frac{21}{29}\right)$(2029,2129) and $\left(\frac{20}{29},\frac{21}{29}\right)$(2029,2129).

Thus, we need to calculate the integral $\int_{-\frac{20}{29}}^{\frac{20}{29}}\ \left(\sqrt{1-x^2}-\frac{21}{29}\right)\ \mathrm{d}x$20292029 (1x22129) dx. As before, we can make this a little simpler by using the symmetry. Thus, 

$\text{area}$area $=$= $2\int_0^{\frac{20}{29}}\ \left(\sqrt{1-x^2}-\frac{21}{29}\right)\ \mathrm{d}x$220290 (1x22129) dx
  $=$= $2\int_0^{\arcsin\frac{20}{29}}\ \left(\cos\theta-\frac{21}{29}\right)\ \cos\theta\mathrm{d}\theta$2arcsin20290 (cosθ2129) cosθdθ
  $=$= $2\int_0^{\arcsin\frac{20}{29}}\ \left(\cos^2\theta-\frac{21}{29}\cos\theta\right)\ \mathrm{d}\theta$2arcsin20290 (cos2θ2129cosθ) dθ
  $=$= $2\int_0^{\arcsin\frac{20}{29}}\ \left(\frac{1}{2}\cos2\theta+\frac{1}{2}-\frac{21}{29}\cos\theta\right)\ \mathrm{d}\theta$2arcsin20290 (12cos2θ+122129cosθ) dθ
  $=$= $2\left[\frac{1}{4}\sin2\theta+\frac{\theta}{2}-\frac{21}{29}\sin\theta\right]_0^{\arcsin\frac{20}{29}}$2[14sin2θ+θ22129sinθ]arcsin20290
  $\approx$ $0.2616$0.2616

We used the same trigonometric substitution and the same identity for $\cos^2\theta$cos2θ as in Example 1.

 

Example 3

 

Find the area enclosed between the graphs of $f(x)=\frac{1}{x}$f(x)=1x and $g(x)=4-x$g(x)=4x, illustrated above.

First, we find the intersection points, where $f(x)=g(x)$f(x)=g(x). That is, 

$\frac{1}{x}$1x $=$= $4-x$4x
$1$1 $=$= $4x-x^2$4xx2
$x^2-4x+1$x24x+1 $=$= $0$0
$(x-2)^2-3$(x2)23 $=$= $0$0
$(x-2+\sqrt{3})(x-2-\sqrt{3})$(x2+3)(x23) $=$= $0$0
$x$x $=$= $2\pm\sqrt{3}$2±3

Now, we calculate $\int_{2-\sqrt{3}}^{2+\sqrt{3}}\ 4-x-\frac{1}{x}\ \mathrm{d}x$2+323 4x1x dx.

This is, 

$\left[4x-\frac{x^2}{2}-\ln x\right]_{2-\sqrt{3}}^{2+\sqrt{3}}$[4xx22lnx]2+323 $\approx$ $4.294$4.294

 

 

Outcomes

M8-11

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

91579

Apply integration methods in solving problems

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