Comprehensive Guide to Surface Area and Volume of Geometric Solids

Surface Area of Circular Solids

• Surface Area of a Half Cylinder:

  • Problem: Find the total surface area (including the rectangular face) of a half cylinder with a radius of $5$ and a height of $2$.

  • Solution Value provided: $35\pi + 20$.

• Painting a Tower Application:

  • The total height of the tower shown is $10\,m$.

  • One liter of paint covers an area of $10\,m^2$.

  • Quantity needed: $24$ cans of 1-L paint are required to paint the entire tower.

• Can Label Dimensions:

  • A can has a diameter of $8\,cm$ and a height of $14\,cm$.

  • Label Size (Length and Width): The label must be $8\pi\,cm$ by $14\,cm$ to fit exactly around the can.

• Cylindrical Wedge Surface Area:

  • A wedge of cheese is cut from a cylindrical block.

  • Find the total surface area: $50\pi + 240$.

Volume of Circular Solids

• Combined Cone and Hemisphere Solid:

  • Given: A solid consisting of a cone and a hemisphere.

  • Total Volume: $324\pi$.

  • Total Surface Area: $117\pi$.

• Ratio of Spheres:

  • Given: Radii of two spheres are in a ratio of $2:3$.

  • Ratio of Volumes: $\frac{2^3}{3^3} = \frac{8}{27}$.

  • Ratio of Surface Areas: $\frac{2^2}{3^2} = \frac{4}{9}$.

• Minisubmarine Dimensions:

  • Total Volume: $138\pi$.

  • Total Surface Area: $105\pi$.

  • Context: Surface area is critical for determining how much pressure the submarine will withstand.

• Inscribed and Circumscribed Spheres in a Cube:

  • The volume of a cube is $1000\,cm^3$.

  • Volume of the largest inscribed sphere: $524\,m^3$ (rounded to the nearest cubic meter).

  • Volume of the smallest circumscribed sphere: $2721\,m^3$ (rounded to the nearest cubic meter).

• Cylindrical Wedge Volume:

  • A wedge of cheese is cut from a cylindrical block.

  • Volume: $200\pi$.

Complex Circular Geometry and Slant Cuts

• Slanted Cylinder Volume:

  • A cylinder is cut on a slant with measurements $6''$ and $8''$.

  • Solid's Volume: $90\pi\,u^3$.

• Right Circular Cone Calculations:

  • Volume: $100\pi\,u^3$.

  • Lateral Area: $65\pi\,u^2$.

  • Total Area: $90\pi\,u^2$.

• Cone with Specific Vertex Angle:

  • Vertex angle: $60^\circ$.

  • Slant height ($l$): $12$.

  • Volume (to nearest tenth): $391.8\,u^3$.

• Rocket Fuel Efficiency Problem:

  • Dimensions provided in diagram.

  • Fuel Requirement: $60\%$ of the space in the rocket is needed for fuel.

  • Available Space: Find the volume available for nonfuel items to the nearest whole unit. Result: $377\,u^3$.

• Ice-Cream Cone Overflow Simulation:

  • Cone measurements: $9\,cm$ deep and $4\,cm$ across the top.

  • Ice cream scoop: $4\,cm$ diameter placed on top.

  • Problem: If the ice cream melts into the cone, will it overflow? Assume volume does not change during melting.

  • Answer: No.

  • Justification: Cone volume ($V_c$) is compared to scoop volume ($V_s$). $V_s \leq V_c$.

Surface Area and Volume of Prisms

• Open Boxes:

  • Calculate the total area of cardboard needed to construct various open boxes.

  • Note: "Open Top" means one face is excluded from the surface area calculation.

• Standard Right Prisms:

  • Right Square Prism: Finding Lateral Area ($LA$) and Surface Area ($SA$).

  • Right Isosceles Triangular Prism: $LA = 360$, $SA = 360 + 60\sqrt{3}$, Volume ($V$) = $300\sqrt{3}$.

• Pentagonal Right Prism:

  • Perimeter of the scalene base: $17$.

  • Lateral edge (height): $10$.

  • Lateral Area: $P \times h = 17 \times 10 = 170$.

Cube Dissection and Probability

• $6$-\text{inch} Cube Problem:

  • A $6-\text{inch}$ cube is painted on the outside and cut into $27$ smaller cubes (each $2-\text{inches}$).

  • Painted faces on small cubes:

    • Six faces painted: $0$ cubes.

    • Five faces painted: $0$ cubes.

    • Four faces painted: $0$ cubes.

    • Three faces painted (Corner cubes): $8$ cubes.

    • Two faces painted (Edge cubes): $12$ cubes.

    • One face painted (Center cubes): $6$ cubes.

    • No faces painted (Interior cube): $1$ cube.

  • Probability: If one small cube is selected at random, the probability it has at least two painted faces is $\frac{20}{27}$.

  • Unpainted Surface Area: Total area of unpainted surfaces is $432\,in^2$.

Practical Applications and Weight

• Waterbed Volume and Weight:

  • Dimensions: $7\,ft$ long, $5\,ft$ wide, $8\,in$ thick (note: $8\,in = \frac{2}{3}\,ft$).

  • Volume: $23.33\,ft^3$ (rounded to nearest cubic foot as $23\,ft^3$).

  • Weight: Water weighs $62.4\,lb/ft^3$.

  • Total weight in Traci's bed: $1456\,lb$.

• Volume-Surface Area Equivalence:

  • Problem: Find the side length of a cube where the numerical value of the volume and surface area are equal.

  • Equation: $s^3 = 6s^2$.

  • Solution: $s = 6$.

Detailed Solid Analysis and Comparisons

• Regular Hexagonal Right Prism:

  • Volume: $720\sqrt{3}$.

  • Surface Area: $240 + 108\sqrt{3}$.

• Hollow Ice Cubes:

  • Dimensions: $4\,cm$ side, hole diameter of $2\,cm$.

  • Volume of ice in one cube: $51.4\,cm^3$.

  • Melting Volume: Water's volume decreases by $11\%$ when changing from solid to liquid. Volume of ten melted cubes = $457.5\,cm^3$.

  • Total Surface Area (including inside hole): $133.4\,cm^2$.

  • Cooling Claim: Manufacturer claims these cool drinks twice as fast as regular cubes. This is evaluated by comparing surface area ratios.

Surface Area and Volume of Pyramids

• Regular Pyramid Properties (PRXYZ):

  • $PR=10$, $RX=12$.

  • Lateral Area: $192$.

  • Square Base Area: $36$ for $ABCD$, $144$ for $RXYZ$.

  • Ratio of areas of bases: $1:4$.

  • Area of trapezoid $ABXR$: $30$.

• Square Pyramid (PABCD):

  • Case 1: Altitude $24$, Slant height $25$.

  • Case 2: Slant height $17$, Altitude $15$.

  • Volume of pyramid with base diagonal $10$ and lateral edge $13$: $\frac{1}{3}Bh$. Volume = $200$.

• Pentagonal Gazebo:

  • Base area: $60\,m^2$.

  • Total height: $16\,m$.

  • Roof height (pyramidal): $6\,m$.

  • Volume Calculation: Volume of prism section ($60 \times 10$) + Volume of pyramid section ($\frac{1}{3} \times 60 \times 6$).

  • Total Volume: $720\,m^3$.

• Regular Tetrahedron:

  • Composed of four equilateral triangles.

  • Volume formula for edge $s$: $V = \frac{s^3}{6\sqrt{2}}$ or $V = \frac{s^3\sqrt{2}}{12}$.

Frustums

• Cone Frustum:

  • Top radius ($r_1$) = $4$, bottom radius ($r_2$) = $8$, slant height ($l$) = $5$.

  • Lateral Area formula: $LA = \pi l (r_1 + r_2) = 60\pi$.

  • Total Surface Area: $60\pi + 16\pi + 64\pi = 140\pi$.

• Pyramidal Frustum:

  • Height: $4\,ft$.

  • Surface Area and Volume calculated based on base areas $B_1$ and $B_2$.

Solids of Rotation

• Rotation about Y-Axis:

  • Region bounded by x-axis, y-axis, $x=7$, and $y=3$ forms a cylinder.

  • $SA = 140\pi$.

  • $Vol = 147\pi$.

• Rotation about X-Axis:

  • Region bounded by $x=-1$, $x=4$, $y=5$, $y=2$.

  • $SA = 112\pi$.

  • $Vol = 105\pi$.

• Complex Rotations:

  • Rotation of $y = -\frac{5}{4}x + 4$ about the y-axis forms a cone. $SA = 36\pi$, $Vol = 12\pi$.

  • Rotation about $y=1$ for $y=-2$, $x=3$, $y=\frac{1}{2}x$.

  • Surface Area results: $24\sqrt{13}\pi\,u^2$ and $96\pi\,u^2$.

Comprehensive Review Data

• Review Set #1 Tasks:

  • Right cylinder $LA=80\pi$, $SA=98\pi$, $Volume=90\pi$.

  • Pyramid with $LA=768$, $SA=912$, $Volume=940.8 + 1728$.

  • Regular hexagonal prism $Volume=240\sqrt{3}$, $SA=216+108\sqrt{3}$.

• Review Set #2 Tasks:

  • Sphere Volume: If $SA = 36\pi$, then $4\pi r^2 = 36\pi \rightarrow r=3$. $Volume = \frac{4}{3}\pi(3)^3 = 36\pi$.

  • Pyramid Volume calculation with base diagonals $7$ and $6$, height $5$: $V=35$.

• Review Set #3 Tasks:

  • Container volumes from folding diagrams.

  • Concrete staircase calculation: Steps $15\,cm$ high, $25\,cm$ deep, $1\,m$ wide with square top platform. Total volume of concrete: $0.5625\,m^3$.

  • Ice cream straw physics: Height $15\,cm$, $17\,cm$ straw fits diagonally. This determines the cylinder's diameter via Pythagorean theorem: $d^2 + 15^2 = 17^2$, so $d=8$.

Questions & Discussion

• Cooling Speeds (Page 6, Problem 13d):

  • Question: Does the hollow ice cube cool a drink twice as fast as a solid cube?

  • Response: This is verified by comparing the Surface Area to Volume ratios. If the ratio of surface area of the hollow cube to the solid cube is $2:1$, the claim is valid because cooling speed is proportional to the surface area exposed to the liquid.

• Ice Cream Melt (Page 3, Problem 11):

  • Question: Will it overflow?

  • Response: No, because the volume of the sphere ($rac{4}{3}\pi r^3$) is less than or equal to the volume of the cone ($rac{1}{3}\pi r^2 h$). Given $r=2$ for both, $V_s = \frac{32}{3}\pi$ and $V_c = \frac{1}{3}\pi(4)(9) = 12\pi$. Since $10.67\pi < 12\pi$, it fits.