1. Stating the problem: We need to find the formulas for surface area and volume for each of the six given shapes composed of cylinders and cones with given height and radius parameters. 2. Important formulas and rules: - Cylinder surface area: $SA_{cyl} = 2\pi r h + 2\pi r^2$ (lateral area plus two bases) - Cylinder volume: $V_{cyl} = \pi r^2 h$ - Cone surface area (lateral only): $SA_{cone} = \pi r s$ where $s = \sqrt{r^2 + h^2}$ is the slant height - Cone volume: $V_{cone} = \frac{1}{3} \pi r^2 h$ - For combined shapes, sum the lateral areas and volumes, subtract base areas if bases are joined. 3. Shape a (cylinder height $l$, radius $r$; cone height $l$, radius $r$): - Surface area: cylinder lateral $2\pi r l$ + cone lateral $\pi r \sqrt{r^2 + l^2}$ + base area $\pi r^2$ (bottom only, top base covered by cone) $$SA_a = 2\pi r l + \pi r \sqrt{r^2 + l^2} + \pi r^2$$ - Volume: $$V_a = \pi r^2 l + \frac{1}{3} \pi r^2 l = \pi r^2 l \left(1 + \frac{1}{3}\right) = \frac{4}{3} \pi r^2 l$$ 4. Shape b (cylinder height $l$, radius $l$; cone height $r$, radius $l$): - Surface area: $$SA_b = 2\pi l l + \pi l \sqrt{l^2 + r^2} + \pi l^2 = 2\pi l^2 + \pi l \sqrt{l^2 + r^2} + \pi l^2 = 3\pi l^2 + \pi l \sqrt{l^2 + r^2}$$ - Volume: $$V_b = \pi l^2 l + \frac{1}{3} \pi l^2 r = \pi l^3 + \frac{1}{3} \pi l^2 r$$ 5. Shape c (two cones joined at bases, each height $l$, radius $l$): - Surface area: sum lateral areas of two cones (no base areas since bases joined) $$SA_c = 2 \times \pi l \sqrt{l^2 + l^2} = 2 \pi l \sqrt{2 l^2} = 2 \pi l^2 \sqrt{2}$$ - Volume: sum volumes of two cones $$V_c = 2 \times \frac{1}{3} \pi l^2 l = \frac{2}{3} \pi l^3$$ 6. Shape d (cylinder height $t$, radius $r$): - Surface area: $$SA_d = 2\pi r t + 2\pi r^2$$ - Volume: $$V_d = \pi r^2 t$$ 7. Shape e (cylinder height $l$, radius $l$; cone height $l$, radius $l$): - Surface area: $$SA_e = 2\pi l l + \pi l \sqrt{l^2 + l^2} + \pi l^2 = 2\pi l^2 + \pi l^2 \sqrt{2} + \pi l^2 = 3\pi l^2 + \pi l^2 \sqrt{2}$$ - Volume: $$V_e = \pi l^2 l + \frac{1}{3} \pi l^2 l = \frac{4}{3} \pi l^3$$ 8. Shape f (oval-like capsule with length parameters $l$, $l$, $l$): - This shape can be approximated as a cylinder of height $l$ and radius $l$ capped by two hemispheres of radius $l$. - Surface area: Cylinder lateral area: $2\pi l l = 2\pi l^2$ Two hemispheres surface area equals one sphere surface area: $4\pi l^2$ $$SA_f = 2\pi l^2 + 4\pi l^2 = 6\pi l^2$$ - Volume: Cylinder volume: $\pi l^2 l = \pi l^3$ Two hemispheres volume equals one sphere volume: $\frac{4}{3} \pi l^3$ $$V_f = \pi l^3 + \frac{4}{3} \pi l^3 = \frac{7}{3} \pi l^3$$ Final answers: - a: $SA = 2\pi r l + \pi r \sqrt{r^2 + l^2} + \pi r^2$, $V = \frac{4}{3} \pi r^2 l$ - b: $SA = 3\pi l^2 + \pi l \sqrt{l^2 + r^2}$, $V = \pi l^3 + \frac{1}{3} \pi l^2 r$ - c: $SA = 2 \pi l^2 \sqrt{2}$, $V = \frac{2}{3} \pi l^3$ - d: $SA = 2\pi r t + 2\pi r^2$, $V = \pi r^2 t$ - e: $SA = 3\pi l^2 + \pi l^2 \sqrt{2}$, $V = \frac{4}{3} \pi l^3$ - f: $SA = 6\pi l^2$, $V = \frac{7}{3} \pi l^3$