find the area enclosed by ellipse x=2cos(theta) and y=3sin(theta). I already kno

Question

find the area enclosed by ellipse x=2cos(theta) and y=3sin(theta). I already know the answer but my question is this,

I set up an equation for r vector r=2cos(theta)i + 3sin(theta)j and took the absolute value then squared it in order touse the formula dA=1/2 r^2d(theta), and I got an aswer of 6.5pi. My book onthe other hand did it by using y and x subbing -2sin(theta)d(theta) for dx and itegrating from -pi/2 to 0. I understand all of that but how come the book gets 6pi and i am getting 6.5pi. Ifeel like bothmethods are valid and should yield the same answer. If someone could clear this up for me I would greatly appreciate it. Thanks inadvance.

Explanation / Answer

x= 2 cos(theta)

x/2 = cos(theta) ....(i)

y = 3 sin(theta)

y/3 = sin (theta) ....(ii)

Squaring (i) and (ii) and adding

x^2/4 +y^2/9 = 1 .............(iii)

y^2/9 = 1- x^2/4

Taking sqrt

y/b = (1/2){4 -x^2}^1/2

y = (3/2) {4 - x^2)^1/2 ............(iv)

since the ellipse is symmetrical about x-axis and y-axis therefore area

=4*integral [y*dx]

= 4* integral [(3/2) (4 - x^2)^1/2 * dx]

= 4*[ 6*{pi/4}]

= 6*pi

I don't know how you have approached (means steps). I can help you more if you provide your steps.

Please rate this. Thanks.

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