Area of an EllipseDate: 08/03/2001 at 23:00:19 From: Tate Subject: Proof I have been given a problem to be solved via integration. I have to show that the area enclosed by the ellipse x^2/a^2 + y^2/b^2 = 1 is pi*a*b, where a is the longer horizontal part and b is the shorter vertical part of an ellipse. Can you offer some suggestions? Date: 08/05/2001 at 03:36:51 From: Doctor Jeremiah Subject: Re: Proof Hi Tate, This ellipse is centered on (0,0), so all we have to do is solve the equation for y to get: x^2/a^2 + y^2/b^2 = 1 b^2*x^2/a^2 + b^2*y^2/b^2 = b^2*1 b^2*x^2/a^2 + y^2 = b^2 b^2*x^2/a^2 + y^2 - b^2*x^2/a^2 = b^2 - b^2*x^2/a^2 y^2 = b^2 - b^2*x^2/a^2 y^2 = b^2/a^2(a^2 - x^2) sqrt( y^2 ) = sqrt( b^2/a^2(a^2 - x^2) ) y = +- b/a * sqrt(a^2 - x^2) This is two equations. One has a plus sign and the other has a minus sign. If we write them as two equations we get: y = b/a * sqrt(a^2 - x^2) y = -b/a * sqrt(a^2 - x^2) The one without the minus sign is the part above the x-axis, and the one with the minus sign is the part below the x-axis. To calculate area between the curve and the x-axis we need to integrate one equation, but to calculate the total area (above and below the x-axis) we need to integrate both equations and then add the areas. But both equations have the same curve and the same area, so we could just integrate one equation and multiply the area by two. If we do that we end up with: / A = 2 * | -b/a * sqrt(a^2 - x^2) dx / Try solving that, and if you get stuck again please write back. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ Date: 08/05/2001 at 10:04:36 From: Tate Subject: Re: Proof To solve / A = 2 * | -b/a * sqrt(a^2 - x^2) dx / I began by pulling out -b/a, as these are constants. The remaining sqrt(a^2 - x^2) dx looks like a good candidate for trig substitution, so I let u = x = a sin theta. Then, my du = a Cos theta. Here is where the train wreck begins. / A = -2(b/a) * | sqrt(a^2 -(a^2 sin^2 theta)) a Cos theta d(theta) / Where do I go now? If I change the Sin^2 to (1-Cos^2), I run into a dead end. Same if I use the double angle identity. My proof is suppose to end up with pi*a*b. How am I going to end up with pi in my solution with a definate integral? If limits are introduced at some point, how do they come to be? Date: 08/10/2001 at 08:48:59 From: Doctor Jeremiah Subject: Re: Proof Hi Tate, First let me say that when I picked the equation to make the integral out of, I should have picked the positive one: y = b/a * sqrt(a^2 - x^2) I picked the negative one, which means the area will be negative. Oops. So I am going to change that now. The equation for the area will be: y = b/a * sqrt(a^2 - x^2) <== one above the x-axis x=a / A = 2 * | b/a * sqrt(a^2 - x^2) dx / x=-a Notice that it's a positive integral now. Your idea of the trig substitution was exactly the right thing to do, but you need to substitute so that the entire square root disappears. I think I would solve this integral like this: The limits on the integral must be -a to a because in the equation x^2/a^2 + y^2/b^2 = 1 the maximum and minimum values of the ellipse occur when it touches the x-axis. When it touches the x-axis, the value of y is 0. When y = 0, then x = a or x = -a depending on the square root. x=a / A = 2 * | b/a * sqrt(a^2 - x^2) dx / x=-a let x = a*sin(t) When we do this our limits of x = a and x = -a become: a = a*sin(t) -a = a*sin(t) 1 = sin(t) -1 = sin(t) Pi/2 = t -Pi/2 = t But more importantly the contents of our square root become: let x = a*sin(t) then x^2 = a^2*sin^2(t) and a^2 - x^2 = a^2 - a^2*sin^2(t) = a^2(1-sin^2(t)) Now you probably remember that sin^2(t) + cos^2(t) = 1. That means that cos^2(t) = 1 - sin^2(t), so a^2 - x^2 = a^2(1-sin^2(t)) = a^2*cos^2(t) and sqrt(a^2 - x^2) = sqrt( a^2*cos^2(t) ) = a*cos(t) Now since x = a*cos(t), then dx must be dx/dt = d/dt( a*sin(t) ) = a*cos(t)*dt Plugging these into the integral we get: t=Pi/2 / A = 2 * | b/a * sqrt(a^2 - x^2) dx / t=-Pi/2 t=Pi/2 / A = 2 * | b/a * a*cos(t) * a*cos(t)*dt / t=-Pi/2 t=Pi/2 / A = 2ba | cos^2(t) dt / t=-Pi/2 Now the easiest way to solve the integral of cos^2(t) is to look it up in a book, but I don't have my book handy. So the integral of cos^2(t) is: By parts: / / | cos^2(t) dt = cos(t)*sin(t) - | sin(t) * -sin(t) dt / / u = cos(t) dv = cos(t)*dt du = -sin(t)*dt v = sin(t) / / | cos^2(t) dt = cos(t)*sin(t) + | sin^2(t) dt / / And since sin^2(t) + cos^2(t) = 1 then sin^2(t) = 1 - cos^2(t) / / | cos^2(t) dt = cos(t)*sin(t) + | 1 - cos^2(t) dt / / / / / | cos^2(t) dt = cos(t)*sin(t) + | dt - | cos^2(t) dt / / / consolidate the cos^2(t) integrals: / / 2 | cos^2(t) dt = cos(t)*sin(t) + | dt / / / 2 | cos^2(t) dt = cos(t)*sin(t) + t / / | cos^2(t) dt = cos(t)*sin(t)/2 + t/2 / Cool, huh? So now we can plug this into the original equation: t=Pi/2 t=Pi/2 / | A = 2ba | cos^2(t) dt = -2ba ( cos(t)*sin(t)/2 + t/2 ) | / | t=-Pi/2 t=-Pi/2 t=Pi/2 | A = 2ba ( cos(t)*sin(t)/2 + t/2 ) | | t=-Pi/2 A = [ 2ba ( cos(Pi/2)*sin(Pi/2)/2 + (Pi/2)/2 ) ] - [ 2ba ( cos(-Pi/2)*sin(-Pi/2)/2 + (-Pi/2)/2 ) ] A = [ 2ba ( (0)*(1)/2 + Pi/4 ) ] - [ 2ba ( (0)*(-1)/2 - Pi/4 ) ] A = [ 2ba*(Pi/4) ] - [ 2ba*(-Pi/4) ] A = [ ba*Pi/2 ] - [ -ba*Pi/2 ] A = ba*Pi/2 + ba*Pi/2 A = ba*Pi And there you are! Let me know if I skimmed over the details too quickly and I will be happy to elaborate on what I did. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/