Question
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!
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!
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