Evaluate $\displaystyle \int_{-8}^8\sqrt{64-x^2}\,dx$

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- Dec 1st 2007, 05:03 PMliyiIntegral with square root
Evaluate $\displaystyle \int_{-8}^8\sqrt{64-x^2}\,dx$

- Dec 1st 2007, 05:22 PMJhevon
the easy way: you could realize that this integral gives you the area of the upper half of the circle with radius 8

the hard way: use the trig substitution $\displaystyle x = 8 \sin \theta$ ....since we have an even function here, you may also wish to change the problem to $\displaystyle 2 \int_0^8 \sqrt{64 - x^2}~dx$, and change the limits based on the trig substitution