AP Calculus BC Practice Quiz: Volumes with Cross Sections: Triangles and Semicircles

Correct Answer: A

The cross sections are perpendicular to the x-axis, so we integrate with respect to x. The diameter of each semicircle at a given x is the distance from the x-axis to the curve y = √x, which is s(x) = √x. The radius is r(x) = (√x)/2. The area of a semicircle is A = (1/2)πr². Substituting r(x), we get A(x) = (1/2)π((√x)/2)² = (1/2)π(x/4) = (π/8)x. The volume is the integral of the area function from x=0 to x=4, which is V = ∫ (π/8)x dx from 0 to 4.

Correct Answer: C

The base of the solid is bounded by y = 4 - x² and y = 0. The intersection points are at x = -2 and x = 2, which are the limits of integration. The cross sections are perpendicular to the x-axis. The side length, s, of each equilateral triangle is the height of the region at a given x, so s(x) = 4 - x². The area of an equilateral triangle with side s is A = (√3/4)s². Therefore, the area of a cross section is A(x) = (√3/4)(4 - x²)². The volume is the integral of this area function: V = ∫ (√3/4)(4 - x²)² dx from -2 to 2.

Correct Answer: B

Cross sections are perpendicular to the y-axis, so we integrate with respect to y. The region is bounded by x = y² and x = 9, which intersect at y = -3 and y = 3. The length of the base of the triangle at a given y is the horizontal distance between the curves: s(y) = (right curve) - (left curve) = 9 - y². Since the cross section is an isosceles right triangle with a leg on the base, its area is A = (1/2)s². So, A(y) = (1/2)(9 - y²)². The volume is V = ∫ (1/2)(9 - y²)² dy from -3 to 3.

Correct Answer: B

This question tests the fundamental concept of calculating volumes with known cross sections. The volume of a solid is found by summing (integrating) the areas of its infinitesimally thin cross sections over a given interval. If the cross-sectional area at a point x is A(x), and the cross sections are taken from x=a to x=b, the total volume is given by the definite integral V = ∫ A(x) dx from a to b. This is the definition of the method of slicing.

Correct Answer: A

The base is bounded by y=cos(x), y=0, and x=0. The region ends where cos(x)=0, which is x=π/2. The cross sections are perpendicular to the x-axis. The diameter of a semicircle at x is s(x) = cos(x). The radius is r(x) = cos(x)/2. The area of a cross section is A(x) = (1/2)πr² = (1/2)π(cos(x)/2)² = (π/8)cos²(x). The volume is V = ∫ (π/8)cos²(x) dx from 0 to π/2. Using the identity cos²(x) = (1+cos(2x))/2, the integral becomes V = (π/8) ∫ (1/2)(1+cos(2x)) dx = (π/16)[x + (1/2)sin(2x)] from 0 to π/2. Evaluating this gives (π/16)[(π/2 + 0) - (0 + 0)] = π²/32. Let me recheck the calculation. V = (π/16)[x + (1/2)sin(2x)] from 0 to π/2. V = (π/16) * [ (π/2 + (1/2)sin(π)) - (0 + (1/2)sin(0)) ] = (π/16) * [π/2] = π²/32. There seems to be an error in my options. Let me correct the options and answer. The correct calculation is π²/32. Let's assume an option was π²/32. Let's re-evaluate the problem to see if I can make it match an option. Ah, let's re-calculate. V = (π/8) ∫ cos²(x) dx from 0 to π/2. ∫ cos²(x) dx = [x/2 + sin(2x)/4]. So V = (π/8) * [x/2 + sin(2x)/4] from 0 to π/2. V = (π/8) * [ (π/4 + sin(π)/4) - (0 + sin(0)/4) ] = (π/8) * (π/4) = π²/32. The options are incorrect. Let me create a new problem or fix the options. Let's fix the options. A: π²/32, B: π/16, C: π/8, D: π²/16. Correct answer is A. Let me re-write the question to fit one of the original options. What if the radius was cos(x)? Then A(x) = (1/2)π(cos(x))². V = (π/2) ∫ cos²(x) dx = (π/2) * (π/4) = π²/8. That's option D. Let's rephrase. 'The radius of the semicircle is given by y=cos(x)'. No, that's not standard. The diameter is the length across the base. The calculation is correct, the options were flawed. I will generate correct options. Let's re-calculate to be absolutely sure. V = ∫ A(x) dx = ∫ (π/8)cos²(x) dx. ∫ cos²(x) dx = ∫ (1+cos(2x))/2 dx = (1/2)x + (1/4)sin(2x). So, V = (π/8) * [(1/2)x + (1/4)sin(2x)] from 0 to π/2. V = (π/8) * [ (π/4 + 0) - (0+0) ] = π²/32. Okay, I will create new options where one is correct. A: π²/32, B: π/16, C: π²/16, D: π/8. The correct answer is A.

Correct Answer: C

First, find the intersection points by setting x = x², which gives x=0 and x=1. These are the limits of integration. The cross sections are perpendicular to the x-axis. The length of the hypotenuse, h, is the vertical distance between the curves: h(x) = (top curve) - (bottom curve) = x - x². For an isosceles right triangle with hypotenuse h, the legs are each h/√2. The area is A = (1/2)(leg)(leg) = (1/2)(h/√2)(h/√2) = (1/2)(h²/2) = h²/4. Substituting h(x), we get A(x) = (1/4)(x - x²)². The volume is V = ∫ (1/4)(x - x²)² dx from 0 to 1.

Correct Answer: B

The cross sections are perpendicular to the x-axis, so we integrate from x=-3 to x=3. For a given x, the base of the equilateral triangle extends from the bottom of the circle (y = -√(9 - x²)) to the top (y = +√(9 - x²)). The side length of the triangle is s(x) = √(9 - x²) - (-√(9 - x²)) = 2√(9 - x²). The area of an equilateral triangle is A = (√3/4)s². Substituting s(x), we get A(x) = (√3/4)(2√(9 - x²))² = (√3/4)(4(9 - x²)) = √3(9 - x²). The volume is the integral of this area: V = ∫ √3(9 - x²) dx from -3 to 3.

Correct Answer: B

The cross sections are perpendicular to the x-axis. The base of each triangle, b, at a given x is the value of the function, so b(x) = e^x. The problem states the height h is half the base, so h(x) = (1/2)b(x) = (1/2)e^x. The area of a triangle is A = (1/2)bh. Substituting the expressions for b(x) and h(x), we get A(x) = (1/2)(e^x)((1/2)e^x) = (1/4)(e^x)² = (1/4)e^(2x). The volume is the definite integral of the area function from x=0 to x=2: V = ∫ (1/4)e^(2x) dx. The setup in option B matches this before simplifying e^(2x).

Correct Answer: A

The cross sections are perpendicular to the y-axis, so we must integrate with respect to y. First, express the boundary curves in terms of y: y = ln(x) becomes x = e^y. The other boundary is x = e. The region is bounded by the y-axis (where y = ln(1) = 0) and the line y = ln(e) = 1. So, the integration is from y=0 to y=1. The diameter of a semicircle at a given height y is the horizontal distance s(y) = (right curve) - (left curve) = e - e^y. The radius is r(y) = (e - e^y)/2. The area of a semicircle is A(y) = (1/2)πr² = (1/2)π((e - e^y)/2)² = (π/8)(e - e^y)². The volume is V = ∫ (π/8)(e - e^y)² dy from 0 to 1.