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Evaluating a definite integral via a quarter-circle area

Consider the function f(x)=16−x2f(x) = \sqrt{16 - x^2}.

The graph of y=f(x)=16−x2y = f(x) = \sqrt{16 - x^2} on the interval [0,4][0,4] is a quarter of a circle centered at the origin.

Evaluating a definite integral via a quarter-circle area graph

Using the fact that this region is a quarter of a circle, evaluate ∫0416−x2 dx.\int_0^4 \sqrt{16 - x^2}\, dx.

Hint

Identify the radius of the full circle from the equation y=16−x2y = \sqrt{16-x^2} (think x2+y2=16x^2+y^2=16), find the area of the full circle, then take the appropriate fraction corresponding to the region shown.

Step-by-step solution

The equation y=16−x2y = \sqrt{16-x^2} describes the upper half of the circle x2+y2=16x^2 + y^2 = 16, which has radius r=4r = 4. Restricting to x∈[0,4]x \in [0,4] picks out exactly the portion of that upper half where x≥0x \ge 0, which is one quarter of the full disk.

The area of a full circle of radius 4 is πr2=16π\pi r^2 = 16\pi. Since the integral represents the area of one quarter of this circle, the value of the integral is 14(16π)=4π\frac{1}{4}(16\pi) = 4\pi.

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