Triple integration with cylindrical coordinates - understanding the intervals

Question

I'm looking at old exams from my school on multivariable calculus, and found the following question:

The paraboloid $z =x^2 + y^2$ and the plane $z = 2y$ encloses a solid. Calculate the volume of K.

I set the equations equal to each other and got $x^2 + (y-1)^2 = 1$, which is a circle of radius 1 around $(0,1)$ on the XY plane. I'm just having trouble understanding how that is translated to the intervals in the triple integral.

This is the correct solution:

$\int_0^\pi\int_0^{2 sin (\theta)}\int_{r^2}^{2rsin(\theta)} r \,dz \,dr \, d\theta = \frac{\pi}{2}$

I can understand the inner interval, it describes how the solid is between the two functions. The middle one (dr) is confusing to me. Isn't that the radius? Shouldn't that simply be 0 to 1? The outer interval is also confusing to me. Shouldn't that be $0$ to $2\pi$ rather than $0$ to $\pi$? I don't understand how this is not a half-circle.

pdf with a solution (Example 42.5 in the document)

Wolfram alpha query for the correct integral

Keep in mind I may be misusing a term somewhere, my education is not in English.

Answer 1

Since the circle is centered at $(0,1)$ we can also use the parametrization

• $0\le \theta \le \pi$
• $0\le r \le 2\sin \theta$

indeed

$$x^2 + (y-1)^2 = 1 \implies r^2\cos^2 \theta +r^2\sin^2\theta-2r\sin \theta=0$$

$$r^2-2r\sin \theta=0 \implies r-2\sin \theta=0 \implies r_{max}=2\sin \theta$$