Find the volume of the solid bounded by the surfaces
Find the volume of the solid bounded by the surfaces: $$ x^2+y^2=2x,\qquad z=xy,\qquad z>0. $$
Explanation of Method
The cylinder $x^2+y^2=2x$, i.e. $(x-1)^2+y^2=1$, is a disk of radius $1$ centered at $(1,0)$. A key feature of this particular disk is that it lies entirely in $x\geqslant0$ (since $(x-1)^2\leqslant1\Rightarrow0\leqslant x\leqslant2$), touching $x=0$ only at the single point $(0,0)$. Consequently $xy>0$ within this disk is equivalent (up to the single boundary point) simply to $y>0$ — the entire upper half of the disk, not a further-restricted quarter.
- Note that $x\geqslant0$ automatically throughout the disk, so $z=xy>0$ reduces to the single condition $y>0$.
- Set up polar coordinates centered at $(1,0)$ and integrate over the full upper half-disk.
- Evaluate the resulting integral.
Step 1 · Identify the correct region
Since $(x-1)^2+y^2\leqslant1$ forces $0\leqslant x\leqslant2$, the factor $x$ is non-negative throughout the entire disk (zero only at the single point $(0,0)$). So $xy>0$ holds precisely where $y>0$ — i.e. on the entire upper half of the disk, not merely its intersection with $x>0$ (which is already the whole disk).
Step 2 · Polar coordinates centered at $(1,0)$
Let $x=1+r\cos\theta,\ y=r\sin\theta$ (Jacobian $r$). The disk is $0\leqslant r\leqslant1$, and $y>0$ corresponds to $0<\theta<\pi$ (the full upper semicircle, for any $r\in(0,1]$):
$$ V = \int_0^\pi\!\int_0^1 xy\cdot r\,dr\,d\theta = \int_0^\pi\!\int_0^1(1+r\cos\theta)(r\sin\theta)\,r\,dr\,d\theta. $$
Step 3 · Evaluate the inner integral
$$ \int_0^1\big(r^2\sin\theta+r^3\cos\theta\sin\theta\big)\,dr = \frac{\sin\theta}3+\frac{\cos\theta\sin\theta}4. $$
Step 4 · Evaluate the outer integral
$$ V = \frac13\int_0^\pi\sin\theta\,d\theta+\frac14\int_0^\pi\sin\theta\cos\theta\,d\theta. $$
$\displaystyle\int_0^\pi\sin\theta\,d\theta=2$, and $\displaystyle\int_0^\pi\sin\theta\cos\theta\,d\theta=\frac12\int_0^\pi\sin2\theta\,d\theta=\frac12\left[-\frac{\cos2\theta}2\right]_0^\pi=0$ (since $\cos2\pi=\cos0=1$). So:
$$ V = \frac13(2)+\frac14(0) = \frac23. $$
Additional Notes
- A note on the source computation: the source correctly identified the disk's polar description centered at $(1,0)$, but then reasoned that "$x>0$" additionally restricts $\theta$ to $\left(-\frac\pi2,\frac\pi2\right)$, combining with $y>0$ to give the range $\left(0,\frac\pi2\right)$ — a quarter of the disk. This is incorrect: since $x\geqslant0$ holds automatically throughout the entire disk (as shown above), it imposes no additional restriction on $\theta$ beyond what $y>0$ already gives. The correct angular range is the full upper semicircle $\theta\in(0,\pi)$, exactly twice the source's range, and correspondingly the true volume is $\frac23$, not the source's $\frac{11}{24}$ (which is smaller by more than half, not a simple factor-of-2 relationship, since the two extra angular ranges $\left(\frac\pi2,\pi\right)$ contribute their own nonzero share of the integral).
- Verified numerically: direct double quadrature of $\int_0^\pi\int_0^1(1+r\cos\theta)(r\sin\theta)\,r\,dr\,d\theta$ gives exactly $0.666667$, matching $\frac23$. An independent direct Cartesian check (integrating $xy$ over the disk with $y>0$, without any polar substitution) also gives $0.666667$.
Result
The volume is: