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: 35π+2035\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 8π cm8\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: 50π+24050\pi + 240.

Volume of Circular Solids

  • Combined Cone and Hemisphere Solid:

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

    • Total Volume: 324π324\pi.

    • Total Surface Area: 117π117\pi.

  • Ratio of Spheres:

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

    • Ratio of Volumes: 2333=827\frac{2^3}{3^3} = \frac{8}{27}.

    • Ratio of Surface Areas: 2232=49\frac{2^2}{3^2} = \frac{4}{9}.

  • Minisubmarine Dimensions:

    • Total Volume: 138π138\pi.

    • Total Surface Area: 105π105\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: 200π200\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: 90π u390\pi\,u^3.

  • Right Circular Cone Calculations:

    • Volume: 100π u3100\pi\,u^3.

    • Lateral Area: 65π u265\pi\,u^2.

    • Total Area: 90π u290\pi\,u^2.

  • Cone with Specific Vertex Angle:

    • Vertex angle: 60∘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 ($V_s$). Vs≤VcV_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+603SA = 360 + 60\sqrt{3}, Volume ($V$) = 3003300\sqrt{3}.

  • Pentagonal Right Prism:

    • Perimeter of the scalene base: $17$.

    • Lateral edge (height): $10$.

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

Cube Dissection and Probability

  • $6−inchCubeProblem:</strong></p><ul><li><p>A-\text{inch} Cube Problem:</strong></p><ul><li><p>A6-\text{inch}$ cube is painted on the outside and cut into $27$ smaller cubes (each 2−inches2-\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 2027\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=23 ft8\,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: 7203720\sqrt{3}.

    • Surface Area: 240+1083240 + 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 $13$: 13Bh\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 (60×1060 \times 10) + Volume of pyramid section (13×60×6\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=s362V = \frac{s^3}{6\sqrt{2}} or V=s3212V = \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=πl(r1+r2)=60πLA = \pi l (r_1 + r_2) = 60\pi.

    • Total Surface Area: 60π+16π+64π=140π60\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=140πSA = 140\pi.

    • Vol=147πVol = 147\pi.

  • Rotation about X-Axis:

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

    • SA=112πSA = 112\pi.

    • Vol=105πVol = 105\pi.

  • Complex Rotations:

    • Rotation of y=−54x+4y = -\frac{5}{4}x + 4 about the y-axis forms a cone. SA=36πSA = 36\pi, Vol=12πVol = 12\pi.

    • Rotation about $y=1$ for $y=-2$, $x=3$, y=12xy=\frac{1}{2}x.

    • Surface Area results: 2413π u224\sqrt{13}\pi\,u^2 and 96π u296\pi\,u^2.

Comprehensive Review Data

  • Review Set #1 Tasks:

    • Right cylinder LA=80πLA=80\pi, SA=98πSA=98\pi, Volume=90πVolume=90\pi.

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

    • Regular hexagonal prism Volume=2403Volume=240\sqrt{3}, SA=216+1083SA=216+108\sqrt{3}.

  • Review Set #2 Tasks:

    • Sphere Volume: If SA=36πSA = 36\pi, then 4πr2=36π→r=34\pi r^2 = 36\pi \rightarrow r=3. Volume=43π(3)3=36πVolume = \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 (rac43πr3rac{4}{3}\pi r^3) is less than or equal to the volume of the cone (rac13πr2hrac{1}{3}\pi r^2 h). Given $r=2$ for both, Vs=323πV_s = \frac{32}{3}\pi and Vc=13π(4)(9)=12πV_c = \frac{1}{3}\pi(4)(9) = 12\pi. Since 10.67\pi < 12\pi, it fits.