Find the volume of the solid bounded by the cylinder

Find the volume of the solid bounded by the cylinder: $$ x^2+y^2-2x=0 $$ and the surface: $$ z=x^2y,\qquad z>0. $$

Step 1 · 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 $y>0$ gives $\theta\in(0,\pi)$.

Step 2 · Expand the integrand

$$ z = x^2y = (1+r\cos\theta)^2\,r\sin\theta = r\sin\theta+2r^2\cos\theta\sin\theta+r^3\cos^2\theta\sin\theta. $$

Step 3 · Set up and evaluate the volume integral

$$ V = \int_0^\pi\!\int_0^1\big(r^2\sin\theta+2r^3\cos\theta\sin\theta+r^4\cos^2\theta\sin\theta\big)\,dr\,d\theta. $$

The inner integral evaluates to:

$$ \int_0^1\big(r^2+2r^3\cos\theta+r^4\cos^2\theta\big)\,dr = \frac13+\frac{\cos\theta}2+\frac{\cos^2\theta}5, $$

so:

$$ V = \int_0^\pi\sin\theta\left(\frac13+\frac{\cos\theta}2+\frac{\cos^2\theta}5\right)d\theta = \frac13\int_0^\pi\sin\theta\,d\theta+\frac12\int_0^\pi\sin\theta\cos\theta\,d\theta+\frac15\int_0^\pi\sin\theta\cos^2\theta\,d\theta. $$

Step 4 · Evaluate each piece

$\displaystyle\int_0^\pi\sin\theta\,d\theta=2$; $\displaystyle\int_0^\pi\sin\theta\cos\theta\,d\theta=0$ (by symmetry about $\theta=\frac\pi2$); and with $u=\cos\theta$: $\displaystyle\int_0^\pi\sin\theta\cos^2\theta\,d\theta=\int_{-1}^1u^2\,du=\frac23$. So:

$$ V = \frac13(2)+\frac12(0)+\frac15\left(\frac23\right) = \frac23+\frac2{15} = \frac{10+2}{15} = \frac{12}{15} = \frac45. $$

Additional Notes

• This derivation matches the source exactly — here $\theta\in(0,\pi)$ is genuinely the correct full range (unlike the superficially similar Problem 154, where an extra, unwarranted restriction to $\theta\in\left(0,\frac\pi2\right)$ crept in; in this problem the height depends on $y$ alone via $z=x^2y$, so no such extra restriction arises).
• Verified symbolically: direct evaluation of the nested integral gives exactly $\dfrac45$.