Find the volume of a solid bounded by $x^2+ y^2 - 2y=0$, $z = x^2 + y^2$, $z=0$ - correctness of solution

Question

The task is to find the the volume of a solid bounded by $x^2+ y^2 - 2y=0$, $z = x^2 + y^2$, $z=0$. Please, check whether the following solution is correct.

We will use double integral and polar coordinates. First, we look for bounds for $r, \theta$: $x^2 + y^2 - 2y = 0$, so $(x+\frac{1}{2})^2 + y^2 = \frac{1}{4}$. Pluging in $x=r\cos(\theta)$, $x=r\sin(\theta)$ we get: $r=0$ or $r=2\sin(\theta)$,

so $r \in (0, 2\sin(\theta))$, $\theta \in (0, 2\pi)$.

Therefore the volume we are looking for is equal to: $$ \int \limits_{0}^{2\pi} \int_{0}^{2 \sin(\theta)} r (r^2 cos^2(\theta)+ r^2 sin^2(\theta)) dr d\theta $$

Answer 1

$$x^2+y^2-2y=0$$ can be reformulated as

$$x^2+(y-1)^2=1$$

The angle doesn't go from $0$ to $2\pi$ but rather just $0$ to $\pi$.

$$ \int \limits_{0}^{\pi} \int_{0}^{2 \sin(\theta)} r (r^2 cos^2(\theta)+ r^2 sin^2(\theta)) dr d\theta =\int \limits_{0}^{\pi} \int_{0}^{2 \sin(\theta)} r^3 dr d\theta $$

Answer 2

If you consider the transformation $$ x= r \cos \theta, \quad y = 1+r \sin \theta $$

the volume can be computed as

$$ \int_0^{2 \pi} \int_0^1 \int_0^{1+r^2+2 r \sin \theta} r \,\,dz dr d \theta= \int_0^{2 \pi} \int_0^1 r (1+r^2+2 r \sin \theta) dr d\theta $$