Question

In Exercises 11-16, evaluate the iterated integral by converting to polar coordinates. $$ \int_{0}^{a} \int_{0}^{\sqrt{a^{2}-y^{2}}} y d x d y $$

Short answer

\(\frac{1}{3}a^{3}\)

Step-by-step solution

Convert to polar coordinates

Substitute x = r cos(θ) and y = r sin(θ). But before doing this, the given integral needs to be re-interpreted in terms of polar coordinates. The given inner integral limits run from 0 to \(sqrt{a^{2}-y^{2}}\) which corresponds to the x-limits of a half circle of radius a in the first quadrant. So, the region is a quarter-disk in the first quadrant. The limits of integral after transformation should reflect this. So, the given integral converts to \(\int_{0}^{\frac{\pi}{2}} \int_{0}^{a} r * r sin(θ) dr dθ \), here r runs from 0 to a and (θ) from 0 to π/2.

Evaluating the inner integral

Evaluate the inner integral first, by taking the integral with respect to r, \( \int_{0}^{a} r^{2}\) sin(θ) dr. This gives us \(\frac{1}{3}a^3 sin(θ)\).

Evaluating the outer integral

Substitute the result of the inner integral into the outer integral, yielding \( \int_{0}^{\frac{\pi}{2}} \frac{1}{3}a^{3} sin(θ) dθ \). Evaluating this gives \(\frac{1}{3}a^{3}[-cos(π/2) + cos(0)] = \frac{1}{3}a^{3}\).