Find the volume of the solid bounded by the surfaces
Find the volume of the solid bounded by the surfaces: $$ z(x+y)=ax+by,\qquad z=0,\qquad1<x^2+y^2<4,\qquad x>0,\ y>0, $$ where $a>0,\,b>0$.
Explanation of Method
Solving for $z=\dfrac{ax+by}{x+y}$ and converting to polar coordinates, the $r$-dependence cancels out of $z$ entirely, leaving $z$ as a function of $\theta$ alone. The volume integral separates into a radial factor (over the annulus) and an angular factor, and the angular integral reduces cleanly using the substitution $\theta\to\frac\pi2-\theta$, which swaps $\cos\theta$ and $\sin\theta$.
- Convert to polar coordinates; note that $z$ depends only on $\theta$.
- Separate the volume integral into radial and angular factors.
- Evaluate the angular integral using the symmetry $\theta\leftrightarrow\frac\pi2-\theta$.
Step 1 · Polar coordinates
With $x=r\cos\theta,\,y=r\sin\theta$: $z=\dfrac{ax+by}{x+y}=\dfrac{a\cos\theta+b\sin\theta}{\cos\theta+\sin\theta}$, independent of $r$. The region is $1<r<2$, $0<\theta<\frac\pi2$.
Step 2 · Separate the integral
$$ V=\int_0^{\pi/2}\frac{a\cos\theta+b\sin\theta}{\cos\theta+\sin\theta}\left[\int_1^2r\,dr\right]d\theta = \frac32\int_0^{\pi/2}\frac{a\cos\theta+b\sin\theta}{\cos\theta+\sin\theta}\,d\theta. $$
Step 3 · Evaluate the angular integral
Split into two pieces and let $I=\displaystyle\int_0^{\pi/2}\frac{\cos\theta}{\cos\theta+\sin\theta}\,d\theta$. Substituting $\theta\to\frac\pi2-\theta$ swaps $\cos\theta$ and $\sin\theta$, showing $\displaystyle\int_0^{\pi/2}\frac{\sin\theta}{\cos\theta+\sin\theta}\,d\theta=I$ as well. Adding the two:
$$ 2I = \int_0^{\pi/2}\frac{\cos\theta+\sin\theta}{\cos\theta+\sin\theta}\,d\theta = \int_0^{\pi/2}1\,d\theta = \frac\pi2\quad\Longrightarrow\quad I=\frac\pi4. $$
So $\displaystyle\int_0^{\pi/2}\frac{a\cos\theta+b\sin\theta}{\cos\theta+\sin\theta}\,d\theta = aI+bI = \frac{\pi(a+b)}4$.
Step 4 · Assemble the volume
$$ V = \frac32\cdot\frac{\pi(a+b)}4 = \frac{3\pi(a+b)}8. $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically (e.g. $a=2.3,\,b=1.7$): direct evaluation of both the reduced single integral and a fully independent direct polar double integral give exactly $\dfrac{3\pi(a+b)}8$.
Result
The volume is: