1. **State the problem:** Find the volume of the solid bounded by the sphere $$x^2 + y^2 + z^2 = 4a^2$$ and the cylinder $$x^2 + y^2 - 2ay = 0$$. 2. **Rewrite the cylinder equation:** The cylinder equation can be rewritten by completing the square for $y$: $$x^2 + y^2 - 2ay = 0 \implies x^2 + (y^2 - 2ay + a^2) = a^2 \implies x^2 + (y - a)^2 = a^2$$ This represents a cylinder of radius $a$ centered along the line $y = a$ in the $xy$-plane. 3. **Interpret the region:** The sphere is centered at the origin with radius $2a$. The cylinder is a circular cylinder of radius $a$ shifted up by $a$ in the $y$-direction. 4. **Set up the volume integral:** Use cylindrical coordinates with $x = r\cos\theta$, $y = r\sin\theta$, and $z = z$. The sphere equation becomes: $$r^2 + z^2 = 4a^2 \implies z = \pm \sqrt{4a^2 - r^2}$$ The cylinder equation in cylindrical coordinates: $$x^2 + (y - a)^2 = a^2 \implies r^2 - 2ar\sin\theta + a^2 = a^2 \implies r = 2a\sin\theta$$ Since $r \geq 0$, and $\sin\theta \geq 0$ for $\theta \in [0, \pi]$, the cylinder restricts $r$ to $0 \leq r \leq 2a\sin\theta$ for $\theta \in [0, \pi]$. 5. **Volume integral:** $$V = \int_{\theta=0}^{\pi} \int_{r=0}^{2a\sin\theta} \int_{z=-\sqrt{4a^2 - r^2}}^{\sqrt{4a^2 - r^2}} r \, dz \, dr \, d\theta$$ 6. **Integrate with respect to $z$:** $$\int_{-\sqrt{4a^2 - r^2}}^{\sqrt{4a^2 - r^2}} dz = 2\sqrt{4a^2 - r^2}$$ So, $$V = \int_0^{\pi} \int_0^{2a\sin\theta} 2r \sqrt{4a^2 - r^2} \, dr \, d\theta$$ 7. **Integrate with respect to $r$:** Let $$I(\theta) = \int_0^{2a\sin\theta} r \sqrt{4a^2 - r^2} \, dr$$ Use substitution: Let $$u = 4a^2 - r^2 \implies du = -2r dr \implies -\frac{1}{2} du = r dr$$ Change limits: When $r=0$, $u=4a^2$; When $r=2a\sin\theta$, $u=4a^2 - 4a^2 \sin^2\theta = 4a^2 \cos^2\theta$. Then, $$I(\theta) = \int_{u=4a^2}^{4a^2 \cos^2\theta} -\frac{1}{2} \sqrt{u} \, du = \frac{1}{2} \int_{4a^2 \cos^2\theta}^{4a^2} u^{1/2} \, du$$ Integrate: $$\int u^{1/2} du = \frac{2}{3} u^{3/2}$$ So, $$I(\theta) = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{4a^2 \cos^2\theta}^{4a^2} = \frac{1}{3} \left( (4a^2)^{3/2} - (4a^2 \cos^2\theta)^{3/2} \right)$$ Calculate powers: $$(4a^2)^{3/2} = (2^2 a^2)^{3/2} = 2^3 a^3 = 8a^3$$ $$(4a^2 \cos^2\theta)^{3/2} = 8a^3 \cos^3\theta$$ Therefore, $$I(\theta) = \frac{1}{3} (8a^3 - 8a^3 \cos^3\theta) = \frac{8a^3}{3} (1 - \cos^3\theta)$$ 8. **Substitute back into volume integral:** $$V = 2 \int_0^{\pi} I(\theta) \, d\theta = 2 \int_0^{\pi} \frac{8a^3}{3} (1 - \cos^3\theta) \, d\theta = \frac{16a^3}{3} \int_0^{\pi} (1 - \cos^3\theta) \, d\theta$$ 9. **Evaluate the integral:** $$\int_0^{\pi} 1 \, d\theta = \pi$$ $$\int_0^{\pi} \cos^3\theta \, d\theta = 0$$ (since $\cos^3\theta$ is an odd function about $\pi/2$ over $[0, \pi]$) So, $$\int_0^{\pi} (1 - \cos^3\theta) \, d\theta = \pi - 0 = \pi$$ 10. **Final volume:** $$V = \frac{16a^3}{3} \pi$$ **Answer:** The volume of the solid bounded by the sphere and the cylinder is $$\boxed{\frac{16 \pi a^3}{3}}$$