Question : Find the area bounded by and above the x-axis

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- Dec 14th 2009, 01:08 AMzorroFind the area of the region bounded
Question : Find the area bounded by and above the x-axis

- Dec 14th 2009, 02:34 AMtonio

By symmetry wrt the x-axis, this is just half the area of the canonical ellipse with semi-axis , i.e. , but of course you need to prove this with integral calculus, so:

, so you need the integral , again by symmetry, this time wrt the y-axis.

You may want to try the trigonometric substitution , to make things easier.

Tonio - Dec 20th 2009, 04:19 PMzorroIs this correct?
- Dec 20th 2009, 04:41 PMoblixps
your set up should look like: 2 integral (from 0 to a) of sqrt(b^2 - (b^2)(x^2)/a^2) dx.

you do not need a trig substitution, simply factor out a b^2/a^2 term inside the square root and you will get sqrt[(b^2/a^2) (a^2 - x^2)]. sqrt(b^2/a^2) becomes b/a so you can factor that constant out of the integral. so now the integral should look like 2b/a integral (from 0 to a) of sqrt(a^2 - x^2) dx. now look at the integral, and geometrically it just means the area of a circle of radius "a" and since the limit is from 0 to a, the integral only wants 1/4 of the whole circle. so integral (from 0 to a) of sqrt(a^2 - x^2) dx is just (pi)(a^2)/4 since it's 1/4 of the area of the whole circle. then with the constant outside of the integral earlier, you get (2b/a)((pi)(a^2)/4)) = (pi)(a)(b)/2 and since the area of an ellipse is pi(a)(b), your result makes sense because you only wanted the top half of the ellipse.

- Dec 20th 2009, 05:01 PMzorroplease look at tonio post
- Dec 20th 2009, 05:07 PMoblixps
I wasn't saying that you shouldn't do the trig substitution. I'm just offering an alternative way of doing the problem which may be easier. you can do the problem either way. all that matters is that you get the correct answer.

- Dec 20th 2009, 06:03 PMzorroOk i interpreted it wrongly
- Dec 20th 2009, 06:49 PMKrahl