AP Calculus AB: Volumes with Known Cross Sections (Drill 1)

Question 1. The region in the xy-plane is bounded above by \( y = 2x \) and below by the x-axis on \( [0, 3] \). Cross sections of the solid perpendicular to the x-axis are squares with side length equal to the height of the base. Which integral gives the volume of the solid?

• A) \( \int_0^3 2x \, dx \)
• B) \( \int_0^3 4x \, dx \)
• C) \( \int_0^3 4x^2 \, dx \) ✓
• D) \( \int_0^3 2x^2 \, dx \)

Explanation: Choice C is correct. The side length of each square cross section equals the base height \( s = 2x \). The area of a square is \( s^2 = (2x)^2 = 4x^2 \), so the volume is \( \int_0^3 4x^2 \, dx \). Choice A is incorrect because the student used the side length directly as the area, writing \( 2x \) instead of \( (2x)^2 \), forgetting to square. Choice B is incorrect because the student doubled the side length rather than squaring it, writing \( 2(2x) = 4x \) instead of \( (2x)^2 = 4x^2 \). Choice D is incorrect because the student applied the area formula for an isosceles right triangle with leg \( s \), that formula is \( \dfrac{1}{2}s^2 \), which gives \( \dfrac{1}{2}(2x)^2 = 2x^2 \), rather than the correct square area formula \( s^2 \).

Question 2. The base of a solid is the region bounded by \( y = \sqrt{4 - x^2} \) and the x-axis. Cross sections perpendicular to the x-axis are semicircles with diameter equal to the height of the base. Which integral gives the volume of the solid?

• A) \( \dfrac{\pi}{8} \int_{-2}^{2} (4 - x^2) \, dx \) ✓
• B) \( \dfrac{\pi}{2} \int_{-2}^{2} (4 - x^2) \, dx \)
• C) \( \dfrac{\pi}{4} \int_{-2}^{2} (4 - x^2) \, dx \)
• D) \( \pi \int_{-2}^{2} (4 - x^2) \, dx \)

Explanation: Choice A is correct. The diameter of each semicircular cross section is \( d = \sqrt{4-x^2} \), giving radius \( r = \dfrac{\sqrt{4-x^2}}{2} \). The area of a semicircle is \( \dfrac{1}{2}\pi r^2 = \dfrac{\pi}{2} \cdot \dfrac{4-x^2}{4} = \dfrac{\pi}{8}(4-x^2) \), so the volume is \( \dfrac{\pi}{8}\int_{-2}^{2}(4-x^2)\,dx \). Choice B is incorrect because the student used the diameter as the radius: \( \dfrac{1}{2}\pi(\sqrt{4-x^2})^2 = \dfrac{\pi}{2}(4-x^2) \), off by a factor of 4. Choice C is incorrect because the student correctly halved the diameter to find \( r \) but applied the full circle formula \( \pi r^2 \) without the \( \dfrac{1}{2} \) for the semicircle, getting \( \dfrac{\pi}{4}(4-x^2) \). Choice D is incorrect because the student used the full circle formula with the diameter as the radius, omitting both the \( \dfrac{1}{4} \) from the radius conversion and the \( \dfrac{1}{2} \) for the semicircle.

Question 3. The base of a solid is the region bounded by \( y = 4 - x^2 \) and \( y = 0 \). Cross sections perpendicular to the x-axis are equilateral triangles with one side lying in the base. Which integral gives the volume of the solid?

• A) \( \int_{-2}^{2} (4 - x^2)^2 \, dx \)
• B) \( \dfrac{\sqrt{3}}{2} \int_{-2}^{2} (4 - x^2)^2 \, dx \)
• C) \( \dfrac{\sqrt{3}}{4} \int_{-2}^{2} (4 - x^2) \, dx \)
• D) \( \dfrac{\sqrt{3}}{4} \int_{-2}^{2} (4 - x^2)^2 \, dx \) ✓

Explanation: Choice D is correct. The side length of each equilateral triangle is \( s = 4 - x^2 \). The area of an equilateral triangle is \( \dfrac{\sqrt{3}}{4}s^2 \), so \( A(x) = \dfrac{\sqrt{3}}{4}(4-x^2)^2 \) and the volume is \( \dfrac{\sqrt{3}}{4}\int_{-2}^{2}(4-x^2)^2\,dx \). The limits are \( \pm 2 \) since the base region exists where \( 4 - x^2 \geq 0 \). Choice A is incorrect because the student dropped the \( \dfrac{\sqrt{3}}{4} \) coefficient entirely, using \( s^2 \) as the area as if the cross sections were squares. Choice B is incorrect because the student used \( \dfrac{\sqrt{3}}{2}s^2 \), double the correct coefficient, possibly confusing the formula with that of a different triangle type. Choice C is incorrect because the student correctly recalled the \( \dfrac{\sqrt{3}}{4} \) coefficient but forgot to square the side length, writing \( \dfrac{\sqrt{3}}{4}(4-x^2) \) instead of \( \dfrac{\sqrt{3}}{4}(4-x^2)^2 \).

Question 4. The table below shows selected values of \( h(x) \), the vertical distance between the upper and lower boundaries of a region. Cross sections of the solid perpendicular to the x-axis are squares with side length \( h(x) \).

x	0	1	2	3
h(x)	1	3	2	4

Using a left Riemann sum with three equal subintervals on \( [0, 3] \), approximate the volume of the solid.

• A) \( 6 \)
• B) \( 14 \) ✓
• C) \( 29 \)
• D) \( 9 \)

Explanation: Choice B is correct. For square cross sections, \( A(x) = [h(x)]^2 \). A left Riemann sum with \( \Delta x = 1 \) uses left endpoints \( x = 0, 1, 2 \): \( V \approx [h(0)]^2(1) + [h(1)]^2(1) + [h(2)]^2(1) = 1^2 + 3^2 + 2^2 = 1 + 9 + 4 = 14 \). Choice A is incorrect because the student forgot to square the side lengths, summing \( h(0) + h(1) + h(2) = 1 + 3 + 2 = 6 \), this gives the Riemann sum for area between curves, not volume. Choice C is incorrect because the student used right endpoints and squared: \( [h(1)]^2 + [h(2)]^2 + [h(3)]^2 = 9 + 4 + 16 = 29 \). Choice D is incorrect because the student used right endpoints without squaring: \( h(1) + h(2) + h(3) = 3 + 2 + 4 = 9 \).

Question 5. A solid has a base in the xy-plane bounded by \( y = \cos x \) and \( y = 0 \) on \( \left[0, \dfrac{\pi}{2}\right] \). Three students each compute a volume using this base but with different cross sections perpendicular to the x-axis: Student 1 uses squares, Student 2 uses equilateral triangles, and Student 3 uses semicircles with diameter equal to the base height. Which of the following correctly ranks the three volumes from greatest to least?

• A) Semicircle > Square > Equilateral triangle
• B) Equilateral triangle > Square > Semicircle
• C) Square > Equilateral triangle > Semicircle ✓
• D) Square > Semicircle > Equilateral triangle

Explanation: Choice C is correct. Each volume has the form \( k\int_0^{\pi/2}[\cos x]^2\,dx \), where \( k \) depends on the cross-section type. For squares: \( k = 1 \). For equilateral triangles: \( k = \dfrac{\sqrt{3}}{4} \approx 0.433 \). For semicircles: \( k = \dfrac{\pi}{8} \approx 0.393 \). Since \( 1 > \dfrac{\sqrt{3}}{4} > \dfrac{\pi}{8} \), the ranking is Square > Equilateral triangle > Semicircle. Choice A is incorrect because the student assumed the circular shape produces the largest volume, not recognizing that \( \dfrac{\pi}{8} \approx 0.393 \), which is less than both 1 and \( \dfrac{\sqrt{3}}{4} \). Choice B is incorrect because the student ranked the equilateral triangle above the square, perhaps believing the \( \sqrt{3} \) factor makes the coefficient exceed 1, but \( \dfrac{\sqrt{3}}{4} \approx 0.433 < 1 \). Choice D is incorrect because the student correctly placed the square first but reversed equilateral triangles and semicircles, not recognizing that \( \dfrac{\pi}{8} \approx 0.393 < \dfrac{\sqrt{3}}{4} \approx 0.433 \).