Solid Geometry #2 - Questions and Answers

Question 1 Determine the solid's volume inside each of the two spheres x² + y² + z² + 4x - 2y + 4z + 5 = 0 and x² + y² + z² = 4 Answer Step 1 x² + y² + z² + 4x - 2y + 4z + 5 = 0 (x²+4x) + (y²-2y) + (z²+4z) + 5 = 0 (x² + 4x +4-4) + (y² - 2y + 1-1) + (z² + 4z + 4-4)+5=0 (x² + 4x + 4) - 4 + (y² - 2y +1) - 1 + (z² + 4z + 4) - 4 +5=0 (x + 2)² + (y - 1)² + (z +2)² = 4 This sphere has a radius of 2 and a center at (-2,1,-2) Step 2 I am going to use the following formula : V = π/12 (4R + d) (2R - d)² where d is the separation between the centers of two spheres And R is the radius of the two spheres, that is 2 in this case Link to the proof of this formula is given in comment section. Step 3 Distance between the two centers is d = √((-2)² + 1² + (-2)²) = √(4 + 1 + 4) = √9 = 3 Step 4 Substitute d=3 and R=2, to get the volume V = (π/12)(4 * 2 + 3)(2 * 2 - 3)² ≈ 2.88 Result V≈ 2.88 Question 2 A paper cup has the shape of a cone with height 10 cm and radius 3 cm (at the top). If water is poured into the cup at a rate of 2 cm³/s, how fast is the water level rising when the water is 5 cm deep? Answer 1 Step 1 If the water level is x and the radius of the cone at this height is rₓ, then we can use the properties of similar triangles to write rₓ/3 = x/10 Multiply both sides by 3 rₓ = 3x/10 = 0.3x The volume of the water in the cone is given by V = ⅓πr²h = ⅓π(0.3x)²x Differentiate both sides with respect to t dV/dt = (π/3)0.09 d(x³)/dt Since the water is being poured at 2cm³/s, dV/dt = 2 Use chain rule in the right-hand side 2 = (π/3)0.09 * 3x² dx/dt Solve for dx / dt dx/dt = 2/((π/3)0.093x²) = 2/(π0.09x²) = 200/(9πx²) When the water is 5 cm deep, we have x=5 dx/dt |x=5 = 200/(9π * 5²) ≈ 0.283cm/s Result water level is rising at the rate of ≈ 0.283cm/s Answer 2 Step 1 V=volume of water r=radius of water h=height of water V = (1/3)πr²h Step 2 Find r in terms of h. r/h = 3/10 r = 3h/10 Step 3 Substitute for r. V = (1/3)π(3h/10)²h V = (3/100)πh³ Step 4 Differentiate dV/dt = (9π/100)h² dh/dt Step 5 Substitute given information 2 = (9π/100) * (5)² * dh/dt Step 6 Solve for dh/dt dh/dt = 2 / (9π * 25/100) dh/dt ≈ 2 / (2.25 π) ≈ 0.283 Result dh/dt ≈ 0.283 cm/sec Question 3 You have two solid steel spheres. Sphere 2 has twice the radius of sphere 1. By what factor does the moment of inertia I₂ of sphere 2 exceed the moment of inertia I₁ of sphere 1? Answer 1 Firstly, I ∝ mr² m = (4/3)πr³ * ρ_steel As sphere 2 has twice the radius, same destiny => its mass is greater by a factor of 2³ = 8 The extra mass is also distributed farther from the center (r₂ = 2r₁), thus, I ∝ mr² ⇒ I₂ ∝ (8m₁)(2r₁)² = 32I₁ Answer 2 To solve the problem, we need to find the factor by which the moment of inertia of sphere 2 exceeds the moment of inertia of sphere 1. Let's denote the radius of sphere 1 as r₁ and the radius of sphere 2 as r₂. Given that sphere 2 has twice the radius of sphere 1, we have r₂ = 2r₁. The moment of inertia of a solid sphere is given by the formula: I = (2/5)mr², where m is the mass of the sphere and r is its radius. Since both spheres are made of solid steel, the masses of the spheres will be proportional to their volumes. The volume of a sphere is given by: V = (4/3)πr³. So, the mass m of a sphere will be proportional to r³. Let's assume the mass of sphere 1 is m₁, and the mass of sphere 2 is m₂. We can write the following equation: m₂ / m₁ = (r₂ / r₁)³ So, the mass of sphere 2 is eight times the mass of sphere 1. Now, let's find the ratio of the moments of inertia: I₂ / I₁ = (m₂r₂²) / (m₁r₁²) = (m₂/m₁) * (r₂/r₁)² Substituting m₂ = 8m₁ and r₂ = 2r₁ into the equation, we have: I₂ / I₁ = (8m₁ / m₁) * (2r₁ / r₁)² = 8 * 2² = 8 * 4 = 32. Therefore, the moment of inertia I₂ of sphere 2 exceeds the moment of inertia I₁ of sphere 1 by a factor of 32. Answer 3 We are given two solid steel spheres, where Sphere 2 has twice the radius of Sphere 1. Let's denote the radius of Sphere 1 as r₁ and the radius of Sphere 2 as r₂. The moment of inertia of a solid sphere is given by the formula: I = (2/5)mr², where m is the mass of the sphere and r is its radius. Since both spheres are made of solid steel, their densities are the same. Hence, their masses will be proportional to their volumes. The volume of a sphere is given by: V = (4/3)πr³. Thus, the mass m of a sphere will be proportional to r³. Let's assume the mass of Sphere 1 is m₁, and the mass of Sphere 2 is m₂. We can write the following equation: m₂/m₁ = (r₂/r₁)³ Since Sphere 2 has twice the radius of Sphere 1, we have r₂ = 2r₁. Substituting this into the equation, we get: m₂/m₁ = (2r₁ / r₁)³ = (2)³ = 8. Thus, the mass of Sphere 2 is eight times the mass of Sphere 1. Now, let's find the ratio of the moments of inertia: I₂/I₁ = (m₂r₂²)/(m₁r₁²) = (m₂/m₁) * (r₂/r₁)² Substituting m₂ = 8m₁ and r₂ = 2r₁ into the equation, we have: I₂/I₁ = (8m₁ / m₁) * (2r₁ / r₁)² = 8 * 2² = 8 * 4 = 32. Therefore, the moment of inertia I₂ of Sphere 2 exceeds the moment of inertia I₁ of Sphere 1 by a factor of 32. Answer 4 I = (2/5)mr² where I is the moment of inertia, m is the mass of the sphere, and r is the radius of the sphere. Let's assume the mass of both spheres is the same. Therefore, we can compare the moments of inertia by using the ratio: I₂ / I₁ = (2/5)m(2r₁)² / (2/5)mr₁² Simplifying the equation, we have: I₂ / I₁ = 4r₁² / r₁² I₂ / I₁ = 4 Therefore, the moment of inertia I₂ of sphere 2 exceeds the moment of inertia I₁ of sphere 1 by a factor of 4. Question 4 Find the surface area of the prism. Answer Given that the length, width and height of the prism are 3ft, 2ft and 4ft respectively. We want to find the surface area of the prism. Surface Area of a Prism Formula:Surface Area of a Prism Formula: The surface area of a prism with length l, width w, and height h is given by Surface area = 2(wl + lh + hw) Using the above formula the surface area is Surface area = 2(2 * 3 + 3 * 4 + 4 * 2) = 2(6 + 12 + 8) = 2(26) = 52 ft² Thus, the surface area of the prism is 52 ft². Question 5 The formula for the volume of a cylinder is V = πr²h. What is the radius for a cylinder that has a volume of 160m³ and a height of 8 m? Express your answer in simplest radical form and as a decimal rounded to the nearest tenth. Answer Using the given formula, we solve for r: V = πr²h Divide both sides by πh V/πh = r² Take the square root: r = √(V/πh) Substitute V = 160m³ and h = 8m r = √(160/π(8)) r = √20/π r = √(4*5)/π r = 2√5/√π r ≈ 2.25