Recent content by pfsm01

If the answer for the definite integral is in fact πb-π√(b^2-a^2 ) for limits x=-a to +a, would it be incorrect to back-calculate the answer for the indefinite integral? For instance, bπ((a+x)/2a) - π√((b^2)-(a^2))((a+x)/2a) with limits x=-a to +a resolves into the answer above.

Apparently the answer is πb-π√(b^2-a^2 ) for limits x=-a to +a but not sure how to arrive here

Would I then integrate with respect to dtheta? If I had to evaluate the integral from -a to +a, would I integrate from 0 to theta after the substitution? Thank you for your help

Homework Statement Evaluate the integral: ∫ {√[(a^2)-(x^2)] / (b-x)} dx Homework Equations ∫ u dv = uv - ∫v du The Attempt at a Solution I've tried using integration by parts but it makes the integral even more complex. I also tried using the table of integrals to find a solution to no...