1. **Problem statement:** Find the volume of the solid in the first octant bounded by the coordinate planes ($x=0$, $y=0$, $z=0$), the cylinder $x^2 + y^2 = 4$, and the plane $z + y = 3$. 2. **Understanding the boundaries:** - The first octant means $x \geq 0$, $y \geq 0$, $z \geq 0$. - The cylinder $x^2 + y^2 = 4$ restricts $x$ and $y$ such that $x^2 + y^2 \leq 4$. - The plane $z + y = 3$ can be rewritten as $z = 3 - y$. 3. **Volume setup:** The volume $V$ can be found by integrating $z$ over the region in the $xy$-plane bounded by the cylinder and the coordinate axes: $$V = \iint_R z \, dA = \iint_R (3 - y) \, dA$$ where $R = \{(x,y) \mid x \geq 0, y \geq 0, x^2 + y^2 \leq 4\}$. 4. **Using polar coordinates:** Since the region is a quarter circle of radius 2 in the first quadrant, use polar coordinates: - $x = r \cos \theta$ - $y = r \sin \theta$ - $r$ ranges from $0$ to $2$ - $\theta$ ranges from $0$ to $\frac{\pi}{2}$ The Jacobian determinant is $r$, so $dA = r \, dr \, d\theta$. 5. **Rewrite the integral:** $$V = \int_0^{\pi/2} \int_0^2 (3 - r \sin \theta) r \, dr \, d\theta = \int_0^{\pi/2} \int_0^2 (3r - r^2 \sin \theta) \, dr \, d\theta$$ 6. **Integrate with respect to $r$:** $$\int_0^2 (3r - r^2 \sin \theta) \, dr = \left[ \frac{3r^2}{2} - \frac{r^3}{3} \sin \theta \right]_0^2 = \frac{3 \times 4}{2} - \frac{8}{3} \sin \theta = 6 - \frac{8}{3} \sin \theta$$ 7. **Integrate with respect to $\theta$:** $$V = \int_0^{\pi/2} \left(6 - \frac{8}{3} \sin \theta \right) d\theta = \left[6\theta + \frac{8}{3} \cos \theta \right]_0^{\pi/2} = 6 \times \frac{\pi}{2} + \frac{8}{3} (\cos \frac{\pi}{2} - \cos 0) = 3\pi + \frac{8}{3} (0 - 1) = 3\pi - \frac{8}{3}$$ 8. **Final answer:** $$\boxed{V = 3\pi - \frac{8}{3}}$$