Home / AP Calculus BC 8.10 Volume with Disc Method: Revolving Around Other Axes – FRQs

AP Calculus BC 8.10 Volume with Disc Method: Revolving Around Other Axes - FRQs - Exam Style Questions

No-Calc Question

Region R bounded by y=xe^{x^2}, y=-2x, and x=1
Let \(R\) be the shaded region bounded by \(y=xe^{x^{2}}\), the line \(y=-2x\), and the vertical line \(x=1\).
(a) Find the area of \(R\).
(b) Write, but do not evaluate, an integral expression that gives the volume of the solid generated when \(R\) is rotated about the horizontal line \(y=-2\).
(c) Write, but do not evaluate, an expression involving one or more integrals that gives the perimeter of \(R\).

Most-appropriate topic codes (CED):

TOPIC 8.4: Area Between Curves (in x) — part (a)
TOPIC 8.10: Disc/Washer About Other Axes — part (b)
TOPIC 8.13: Arc Length of Smooth Curves — part (c)

▶️ Answer/Explanation

(a)
Let \(y_1=xe^{x^{2}}\), \(y_2=-2x\). On \(0\le x\le 1\), \(y_1\ge y_2\).
\(\displaystyle \text{Area}=\int_{0}^{1}\big(y_1-y_2\big)\,dx =\int_{0}^{1}\!\big(xe^{x^{2}}+2x\big)\,dx\)
\(=\left[\tfrac12\,e^{x^{2}}+x^{2}\right]_{0}^{1} =\left(\tfrac12 e+1\right)-\left(\tfrac12+0\right) =\boxed{\tfrac12 e+\tfrac12}.\)

(b)
Rotate about \(y=-2\) (washers; vertical slices).
\(R_{\text{out}}(x)=(y_1+2)\), \(R_{\text{in}}(x)=(2-2x)\).
\(\displaystyle \boxed{V=\pi\int_{0}^{1}\Big[(y_1+2)^{2}-(2-2x)^{2}\Big]\,dx}\), where \(y_1=xe^{x^{2}}\).

(c)
Perimeter \(=\) arc of \(y_1\) on \([0,1]\) \(+\) arc of \(y_2\) on \([0,1]\) \(+\) the vertical segment at \(x=1\).
\(y_1′(x)=e^{x^{2}}(1+2x^{2})\), \(y_2′(x)=-2\).
\(\displaystyle \boxed{ \text{Perimeter}= \int_{0}^{1}\!\sqrt{1+\big(e^{x^{2}}(1+2x^{2})\big)^{2}}\,dx +\int_{0}^{1}\!\sqrt{1+(-2)^{2}}\,dx +(e-(-2)) }\)
\(=\;\int_{0}^{1}\!\sqrt{1+\big(e^{x^{2}}(1+2x^{2})\big)^{2}}\,dx +\sqrt{5} × 1 +(e+2).\)

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