Sketch the region given by the definite integral. Use geometric shapes and formulas to evaluate the integral (a > 0, r > 0).
r
∫ sqrt(r^2 - x^2) dx
-r
While I recognize that this looks similar to a circle function, I'm not sure how to graph and evaluate this definite integral because the lower/upper bounds is throwing me off because it's not giving us a number, but a variable.
How could we approach this problem? I know there's tons of calculators that can calculate the integral by finding the anti derivative, but this is not the case for this problem specifically.
Any help is greatly appreciated!
2 answers
Should we assume r = 1?
Do not assume r=1. But, it is a constant. And you are right, the graph is a sem-circle, so the area is π/2 r^2
Not sure where the a comes in, but to do the integral, now is the time to start getting used to trig substitutions. Let
x = r sinθ
r^2-x^2 = r^2(1-sin^2θ) = r^2 cos^2θ
dx = r cosθ dθ
Now you have
∫[-π/2,π/2] r cosθ (r cosθ dθ)
= r^2 ∫[-π/2,π/2] cos^2θ dθ
= r^2/2 ∫[-π/2,π/2] (1+cos2θ) dθ
= r^2/2 (θ + 1/2 sin2θ) [[-π/2,π/2]
= r^2/2 [(π/2+0)-(-π/2+0)]
= πr^2/2
Not sure where the a comes in, but to do the integral, now is the time to start getting used to trig substitutions. Let
x = r sinθ
r^2-x^2 = r^2(1-sin^2θ) = r^2 cos^2θ
dx = r cosθ dθ
Now you have
∫[-π/2,π/2] r cosθ (r cosθ dθ)
= r^2 ∫[-π/2,π/2] cos^2θ dθ
= r^2/2 ∫[-π/2,π/2] (1+cos2θ) dθ
= r^2/2 (θ + 1/2 sin2θ) [[-π/2,π/2]
= r^2/2 [(π/2+0)-(-π/2+0)]
= πr^2/2
5 answers