1. **State the problem:** We need to express the total volume $V$ of the combined cylindrical cans in terms of the radius $r$ of the smaller cylinder. 2. **Recall the volume formula for a cylinder:** $$V = \pi r^2 h$$ where $r$ is the radius and $h$ is the height. 3. **Identify the volumes of each cylinder:** - Small cylinder: radius $r$, height $h$, volume $V_1 = \pi r^2 h$ - Large cylinder: radius $2r$, height $2h$, volume $V_2 = \pi (2r)^2 (2h) = \pi \times 4r^2 \times 2h = 8 \pi r^2 h$ 4. **Total volume:** $$V = V_1 + V_2 = \pi r^2 h + 8 \pi r^2 h = 9 \pi r^2 h$$ 5. **Express $h$ in terms of $r$ using part (a):** From part (a), the surface area of both cylinders is $400\pi$ cm². Surface area of small cylinder: $$A_1 = 2\pi r^2 + 2\pi r h$$ Surface area of large cylinder: $$A_2 = 2\pi (2r)^2 + 2\pi (2r)(2h) = 8\pi r^2 + 8\pi r h$$ Total surface area: $$A = A_1 + A_2 = (2\pi r^2 + 2\pi r h) + (8\pi r^2 + 8\pi r h) = 10\pi r^2 + 10\pi r h = 400\pi$$ Divide both sides by $\pi$: $$10 r^2 + 10 r h = 400$$ Solve for $h$: $$10 r h = 400 - 10 r^2$$ $$h = \frac{400 - 10 r^2}{10 r} = \frac{40 - r^2}{r}$$ 6. **Substitute $h$ into the total volume formula:** $$V = 9 \pi r^2 \times \frac{40 - r^2}{r} = 9 \pi r (40 - r^2) = 9 \pi (40 r - r^3)$$ **Final expression:** $$\boxed{V = 9 \pi (40 r - r^3)}$$ This expresses the total volume $V$ in terms of $r$ only.