AP Calculus BC 8.10 Volume with Disc Method: Revolving Around Other Axes - FRQs - Exam Style Questions
No-Calc Question

Most-appropriate topic codes (CED):
• 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 \(x=1\).
\(y_1′(x)=e^{x^{2}}(1+2x^{2})\), \(y_2′(x)=-2\).
\(\displaystyle \boxed{ \text{Perimeter}= \int_{0}^{1}\!\sqrt{1+\big(e^{x^{2}}(1+2x^{2})\big)^{2}}\,dx +\int_{0}^{1}\!\sqrt{1+(-2)^{2}}\,dx +(e-(-2)) }\)
\(=\;\int_{0}^{1}\!\sqrt{1+\big(e^{x^{2}}(1+2x^{2})\big)^{2}}\,dx +\sqrt{5} × 1 +(e+2).\)