Question 272208
Hi There
For my maths assignment I need to prove that the area of an ellipse is given by the formula {{{A = pi*ab}}}

<pre><font size = 4 color = "indigo"><b>
{{{drawing(400,300,-4,4,-3,3,  line(-4,0,3.5,0), line(0,-3,0,2.6),locate(3.6,.2,x), locate(-.1,2.97,y), locate(0,2.3,"(0,b)"), locate(3,.5,"(a,0)"),
rectangle(1.5,0,1.6,1.7),locate(1.7,.9,y), locate(1.4,0,dx),
arc(0,0,6,-4,0,360)   )}}}

The equation of the above ellipse is 

{{{x^2/a^2+y^2/b^2=1}}}

We will just find the area of the part in the first quadrant,
and then by symmetry multiply by 4 to get the area in all four 
quadrants.  We will use:

{{{A=4int(y, dx, 0, a )}}}


We solve for y:

{{{x^2/a^2+y^2/b^2=1}}}

Multiply through by {{{a^2b^2}}}

{{{b^2x^2+a^2y^2=a^2b^2}}}

{{{a^2y^2=a^2b^2-b^2x^2}}}

{{{a^2y^2=b^2(a^2-x^2)}}}

Take positive square roots of both sides:

{{{ay=b*sqrt(a^2-x^2)}}}

{{{y=(b/a)*sqrt(a^2-x^2)}}}

{{{A=4int((b/a)*sqrt(a^2-x^2), dx, 0, a) =

4(b/a)int(sqrt(a^2-x^2), dx, 0, a)                   }}}

We use trigonometric substitution:
sqrt(a^2-x^2)



{{{drawing(200,200, -1,2,-1,2, triangle(-.5,0,1.5,0,1.5,1.5),
locate(.3,1,a), locate(.2,-.2,sqrt(a^2-x^2)), locate(1.6,.9,x) 
locate(-.2,.3,"@")  )}}}

-----------------------------------


{{{x/a=sin("@")}}}

{{{x=a*sin("@")}}}

{{{dx=a*cos("@")*"d@"}}}

{{{sqrt(a^2-x^2)/a=cos("@")}}}

{{{sqrt(a^2-x^2)=a*cos("@")}}}

To substitute we eill also change the limits from
limits on x to limits on @:

When {{{x=0}}}, {{{"@"=0}}} and
when {{{x=a}}}, {{{"@"=pi/2}}} 


Substituting

{{{A=4int( (b/a)*sqrt(a^2-x^2), dx, 0, a) }}} 

{{{A=4int( (b/a)*a*cos("@")*a*cos("@"), "d@",0, pi/2 ) }}}

{{{A=4int((b/cross(a))*cross(a)*cos("@")*a*cos("@"), "d@",0, pi/2 )  
         }}}

{{{A=4int(ab*Cos^2("@"), "d@",0, pi/2 ) =  4ab*int(Cos^2("@"), "d@",0, pi/2 )
 
=4ab*int((1/2)(1+cos("2@")), "d@",0, pi/2 )

=(4ab/2)*int((1+cos("2@")), "d@",0, pi/2 )   

      }}}

{{{A=(4ab/2)*( int("", "d@") + int(cos("2@"), "d@" ) )matrix(2,2,"|",pi/2,"|",0)=
(4ab/2)*( "@" + (1/2)int(cos("2@"), ""*2*"d@" ) )matrix(2,2,"|",pi/2,"|",0)

  }}}
  
{{{A= 2ab*( "@" + (1/2)(sin("2@"))) matrix(2,2,"|",pi/2,"|",0)

= 2ab((pi/2-0) + (1/2)(sin(2*(pi/2))-sin(2*0)))


 }}}

{{{A=2ab(pi/2 + (1/2)(sin(cross(2)*(pi/cross(2)))-sin(0)))


 }}}

{{{A= 2ab(pi/2 + (1/2)(sin(pi)-sin(0))) }}}

{{{A= 2ab(pi/2 + (1/2)(0-0)) }}}

{{{A= 2ab(pi/2 + 0) }}}

{{{A= 2ab(pi/2) }}}

{{{A= cross(2)ab(pi/cross(2)) }}}

{{{A=pi*ab}}}

Edwin</pre>