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Calculus/The Fundamental Theorem of Calculus

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Expert: - 3/1/2010


Question
Evaluate the integral by interpreting it as area and using formulas from
geometry.

(integral from -8 to 8) of 2 times the square root of (64-x^2) dx.

I know the answer is supposed to be 64pi, but I keep getting 100.53, because the shape is a half circle with a radius of 8.
--> (1/2)(pi)(8)^2 = 100.53

Answer
The problem is really ∫∫ 1 dy dx, where y goes from 0 to √(64-x²) and x goes from -8 to 8.

Convert to polar coordinates and note that ∫∫ f(x,y) dx dy = ∫∫ r*f(r,Θ) dr dΘ.

Doing the conversion, it can be seen that r goes from 0 to 8 and Θ goes from 0 to π.

Also, since f(x,y) = 1, f(r,Θ) = 1.  The problem is 2∫∫ r dr dΘ.

That is 2(r²/2) for r from 0 to 8, which is 64, and so we have ∫32 dΘ.

The integral in terms of Θ is 64Θ evaluated from 0 to π, so the answer is 64π.

This is πr².


This was what you most likely did, and the only place I think there is a mistake is
was forgetting that is said to take 2 * (yes, that's two times) the result.

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