Why does $\int_0^\pi \int_0^1 r^2 $cos$\theta\ dr d \theta \neq 2\int_0^{\pi/2} \int_0^1 r^2 $cos$\theta\ dr d \theta $

Question

Apparently this inequality:

$$\int_0^\pi \int_0^1 r^2 \cos \theta\ dr d \theta \neq 2\int_0^{\pi/2} \int_0^1 r^2 \cos \theta\ dr d \theta $$

Is true. But why? I sketched out the regions of integration and the right hand side is the top half of the unit circle, and the left hand side is simply half of that hemisphere. So multiplying it by two gives the same region, and since the integrands are the same shouldn't these integrals be equal?

Answer 1

The function $\cos \theta $ is positive in the first quadrant but changes sign in the second quadrant so the function values are not symmetrical.

Answer 2

Because $$0=\int_0^\pi \cos\theta\ d\theta\neq2\int_0^{\pi\over2} \cos\theta\ d\theta=2$$

Answer 3

because $r$ and $\theta$ are separable, this is the same as saying: $$\int_0^\pi\cos(\theta) d\theta\ne2\int_0^{\pi/2}\cos(\theta)d\theta$$ and since: $$\int_{\pi/2}^\pi\cos(\theta)d\theta=-\int_0^{\pi/2}\cos(\theta)d\theta$$ $$\int_0^\pi\cos(\theta)d\theta=0$$ whilst: $$\int_0^{\pi/2}\cos(\theta)d\theta\ne0$$