using polar integration to solve $\iint\limits_D \sqrt{R^2-x^2 - y^2} dxdy, \ D: x^2 + y^2 \le Rx$

Question

$$I = \iint\limits_D \sqrt{R^2-x^2 - y^2} dxdy, given\ D: x^2 + y^2 \le Rx$$ solving this using polar integration, $$(r\cos\theta)^2 + (r\sin\theta)^2 = R \cdot r\cos \theta$$ $$ -\frac\pi2 \le \theta \le \frac\pi2, 0 \le r \le R\cos \theta$$ $$I= \int_{-\frac\pi2}^{\frac\pi2} d\theta \int_0^{R\cos\theta}\sqrt{R^2-r^2} r dr =-\frac12\cdot\frac23\int_{-\frac\pi2}^{\frac\pi2} ((R^2-R^2\cos^2\theta)^\frac32-R^3)\ d\theta=$$ $$=-\frac13\int_{-\frac\pi2}^{\frac\pi2}(R^3\sin^3\theta-R^3)\ d\theta =-\frac13{R^3}(-\frac13\sin^2\theta\cos\theta - \frac23\cos\theta-\theta)\big|_{-\frac\pi2}^{\frac\pi2}=\frac\pi3R^3$$

The sine reduction formulae used is $$\int \sin^n x\ dx = −\frac1n \cos x \sin^{n-1} x +\frac{n−1}n \int (\sin^{n-2} x) dx$$ Which step is wrong?

Answer 1

The trap is here : One have to take the absolute value of the sine.

Answer 2

Well, it seems that the problem is in the calculation of the integral. We have (regarding the indefinite integral)$$\int (\sin^3\theta-1)d\theta=\int((1-\cos^2\theta)\sin\theta-1)d\theta=\int(\sin\theta-\cos^2\theta\sin\theta-1)d\theta =-\cos\theta +\frac{1}{3}\cos^3\theta -\theta+c.$$