wilcofan3
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Homework Statement
Find the length of the ellipse 9x^2 + 10y^2 = 90 correct to six decimal places.
Homework Equations
4Larc in the first quadrant = Lellipse
The Attempt at a Solution
Just checking to see if I did this right:
9x^2 + 10y^2 = 90
x^2/10 + y^2/9 = 1
Therefore a = \sqrt{10} and b = 3.
This makes x = \sqrt{10}sin t and y = 3 cos t
Since \int \sqrt{(dx)^2 + (dy)^2} is the formula for arc length, do I just get: 4\int \sqrt{10(cos t)^2 - 9(sin t)^2}?
And are the bounds just from 0 to \pi/2?
Thanks!