Find the volume of the solid bounded by the surfaces
Find the volume of the solid bounded by the surfaces: $$ x^2+y^2=2x,\qquad x^2+y^2=2y,\qquad z=x+2y,\qquad z=0. $$
Explanation of Method
The two cylinders are unit circles $(x-1)^2+y^2=1$ and $x^2+(y-1)^2=1$, centered at $(1,0)$ and $(0,1)$, intersecting at $(0,0)$ and $(1,1)$. The base region $D$ is their lens-shaped intersection (a vesica piscis). Over $D$, the plane $z=x+2y$ gives the height, and the volume is $\iint_D(x+2y)\,dA$.
- Identify the lens region and its bounding arcs for $x\in[0,1]$.
- Set up the volume as an iterated integral.
- Evaluate the inner integral, then integrate the resulting expression over $x$.
Step 1 · The lens region
For $x\in[0,1]$, the intersection of the two disks $(x-1)^2+y^2\leqslant1$ and $x^2+(y-1)^2\leqslant1$ gives $y$ ranging from $y_1=1-\sqrt{1-x^2}$ (the lower arc, from the second circle) to $y_2=\sqrt{1-(x-1)^2}$ (the upper arc, from the first circle) — confirmed by checking that these are indeed the tighter bounds throughout $x\in[0,1]$.
Step 2 · Set up the volume integral
$$ V=\int_0^1\int_{y_1}^{y_2}(x+2y)\,dy\,dx = \int_0^1\Big[x(y_2-y_1)+\big(y_2^2-y_1^2\big)\Big]dx. $$
Step 3 · Expand the integrand
Since $y_2^2=1-(x-1)^2=2x-x^2$ and $y_1^2=1-2\sqrt{1-x^2}+(1-x^2)=2-x^2-2\sqrt{1-x^2}$:
$$ x(y_2-y_1)+\big(y_2^2-y_1^2\big) = x\sqrt{2x-x^2}+x\sqrt{1-x^2}+2\sqrt{1-x^2}+x-2. $$
Step 4 · Integrate term by term
Using $2x-x^2=1-(x-1)^2$ and the substitution $t=x-1$:
$$ \int_0^1x\sqrt{2x-x^2}\,dx = \int_{-1}^0(t+1)\sqrt{1-t^2}\,dt = -\frac13+\frac\pi4. $$
$$ \int_0^1x\sqrt{1-x^2}\,dx=\frac13,\qquad \int_0^12\sqrt{1-x^2}\,dx=\frac\pi2,\qquad \int_0^1(x-2)\,dx=-\frac32. $$
Summing all four pieces:
$$ V = \left(-\frac13+\frac\pi4\right)+\frac13+\frac\pi2-\frac32 = \frac{3\pi}4-\frac32. $$
Additional Notes
- A note on the source computation: the source correctly identified the lens region and set up the same iterated integral, but stopped before completing the evaluation ("this becomes quite involved"), then presented an unjustified final guess of $\dfrac{5\pi-16}6\approx-0.0487$ — notably negative, which is impossible for a volume. A full completion of the source's own (correctly set up) integral, verified independently by direct numerical double integration, gives the positive value $V=\dfrac{3\pi}4-\dfrac32\approx0.8562$.
Result
The volume is: