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2021 AP Calculus AB FRQ Question 3: Spinning Toy — Volume of Revolution

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2021.

Question 3

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Spinning Toy — Volume of Revolution

Area & Volume

Hard
A company designs spinning toys using the family of functions y=cx4−x2y=cx\sqrt{4-x^2}, where cc is a positive constant. The region in the first quadrant bounded by the xx-axis and the graph of y=cx4−x2y=cx\sqrt{4-x^2}, for some cc, is revolved about the xx-axis to form each spinning toy (a solid of revolution). Both xx and yy are measured in inches. It is known that dydx=c(4−2x2)4−x2\dfrac{dy}{dx}=\dfrac{c(4-2x^2)}{\sqrt{4-x^2}}.
Part AMedium2 points
Find the area of the region in the first quadrant bounded by the xx-axis and the graph of y=cx4−x2y=cx\sqrt{4-x^2} for c=6c=6.

Answer

1616 in²
Full Solution & Work

Set up the area integral

Area=∫026x4−x2 dx\text{Area} = \int_0^2 6x\sqrt{4-x^2}\,dx

Substitute u = 4 − x²

du=−2x dxdu=-2x\,dx. When x=0,u=4x=0,u=4; when x=2,u=0x=2,u=0.
=6⋅(−12)∫40u du=3∫04u1/2 du=3[23u3/2]04= 6\cdot\left(-\frac12\right)\int_4^0 \sqrt u\,du = 3\int_0^4 u^{1/2}\,du = 3\left[\frac23 u^{3/2}\right]_0^4

Evaluate

=2(43/2)=2(8)=16 in2= 2\big(4^{3/2}\big) = 2(8)=16 \text{ in}^2

AP Scoring — 2 Points

**P1**: Correct integral setup with the substitution u=4−x2u=4-x^2.
**P2**: Correct answer, 16.
Part BMedium2 points
For a particular spinning toy, the radius of the largest cross-sectional circular slice is 1.2 inches. What is the value of cc for this spinning toy?

Answer

c=0.6c=0.6
Full Solution & Work

Find where y is maximized

dydx=0  ⟹  4−2x2=0  ⟹  x=2\frac{dy}{dx}=0 \implies 4-2x^2=0 \implies x=\sqrt2 (the only critical point in [0,2][0,2]).

Find the maximum radius in terms of c

y(2)=c2⋅4−2=c2⋅2=2cy\big(\sqrt2\big) = c\sqrt2\cdot\sqrt{4-2} = c\sqrt2\cdot\sqrt2=2c

Solve for c

2c=1.2  ⟹  c=0.62c = 1.2 \implies c=0.6

AP Scoring — 2 Points

**P1**: Correctly finds the maximum occurs at x=2x=\sqrt2, with the largest radius equal to 2c2c.
**P2**: Correct answer, c=0.6c=0.6.
Part CHard3 points
For another spinning toy, the volume is 2π2\pi cubic inches. What is the value of cc for this spinning toy?

Answer

c=1532=308≈0.685c=\sqrt{\dfrac{15}{32}}=\dfrac{\sqrt{30}}{8}\approx0.685
Full Solution & Work

Set up the volume of revolution

V=π∫02(cx4−x2)2 dx=πc2∫02x2(4−x2) dxV = \pi\int_0^2 \big(cx\sqrt{4-x^2}\big)^2\,dx = \pi c^2\int_0^2 x^2(4-x^2)\,dx

Evaluate the integral

∫02(4x2−x4) dx=[4x33−x55]02=323−325=6415\int_0^2 (4x^2-x^4)\,dx = \left[\frac{4x^3}{3}-\frac{x^5}{5}\right]_0^2 = \frac{32}{3}-\frac{32}{5} = \frac{64}{15}

Solve for c

πc2⋅6415=2π  ⟹  c2=3064=1532  ⟹  c=1532=308≈0.685\pi c^2\cdot\frac{64}{15} = 2\pi \implies c^2 = \frac{30}{64}=\frac{15}{32} \implies c=\sqrt{\frac{15}{32}}=\frac{\sqrt{30}}{8}\approx0.685

AP Scoring — 3 Points

**P1**: Correct volume-of-revolution setup, π∫02(cx4−x2)2 dx\pi\int_0^2(cx\sqrt{4-x^2})^2\,dx.
**P2**: Correctly evaluates ∫02x2(4−x2) dx=6415\int_0^2 x^2(4-x^2)\,dx=\frac{64}{15}.
**P3**: Correct final answer, c=308≈0.685c=\frac{\sqrt{30}}{8}\approx0.685.

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