Find the area enclosed?

Question

Find the area enclosed by the ellipse with parametric equations x=2 cos θ and y=3 sin θ. 

 

Answer: 6pi

 

Please show all your work.

Answer 1

Sun,

 

The key to find the area of an ellipse is to find the length of its axes.  All of the x-coordinates are governed by cosine, which is at is maximum at θ=0 and at it minimum at θ=pi.  x must also be zero at ±pi/2.  So converse is true for sine (which governs y), which is at is maximum at pi/2; its minimum at -pi/2; and zero at 0 and pi.  So this ellipse has the ends of its x coordinates at (-2,0) and (2,0), and the ends of its y coordinates at (0,-3) and (0,3).  The formula for the area of the ellipse is pi x a x b where a and b are the semi-lengths of the axes or pi x 2 x 3 = 6pi.  If you wanted to convert this into a more standard equation for an ellipse, note that x/2=cos and y/3=sin, so (x/2)^2+(y/3)^2= cos^2 + sin^2 = 1 or

(x^2/2^2) + (y^2/3^2) = 1, which is the standard equation for an ellipse.

 

Unfortunately, this probably isn't what your instructor is looking for.  Take a look at

 

http://tutorial.math.lamar.edu/Classes/CalcII/ParaArea.aspx

 

for a complete description.  But basically, you need to integrate

2pi

∫y(θ)x'(θ) dθ or conversely ∫x(θ)y'(θ) dθ (over the same interval)

0

 

Given, x=2 cos θ and y=3 sin θ, dx/dθ= -2 sinθ so we need to integrate ∫3 sin θ (-2 sin θ) dθ= -6 ∫ sin^2θ dθ = -6 [x/2 - 1/4(sin2x)] at 2pi and 0.  If you evaluate this you will get -6pi, but this indicates that integral should have been from 2pi to 0 instead, if we wanted positive area.  The adjustment of the limits matters more if the problem involved less than (or more than) rotation around the origin (which the web page discusses that adjustment in more detail).