AP Calculus AB 8.8 Volumes with Cross Sections: Triangles and Semicircles - MCQs - Exam Style Questions
Calc-Ok Question
The base of a solid is a right triangle with legs \(6\) and \(16\). The hypotenuse is \(y=-\tfrac{3}{8}(x-16)\) on \(0\le x\le 16\). Each cross section perpendicular to the \(x\)-axis is a semicircle. What is the volume of the solid?
(A) \(75.398\)
(B) \(150.796\)
(C) \(301.593\)
(D) \(603.186\)
(B) \(150.796\)
(C) \(301.593\)
(D) \(603.186\)
▶️ Answer/Explanation
For \(0\le x\le 16\), height \(y=6-\frac{3}{8}x\).
Semicircle radius \(r=\dfrac{y}{2}=3-\dfrac{3x}{16}\).
Area \(A(x)=\dfrac{\pi r^{2}}{2}=\dfrac{\pi}{2}\Big(3-\dfrac{3x}{16}\Big)^{2}\).
Volume \(\displaystyle V=\int_{0}^{16}A(x)\,dx=\frac{\pi}{2}\int_{0}^{16}\!\Big(3-\frac{3x}{16}\Big)^{2}dx =\frac{\pi}{2}\Big[9x-\frac{9}{16}x^{2}+\frac{9}{768}x^{3}\Big]_{0}^{16}=24\pi\approx 75.398.\)
✅ Answer: (A)
Semicircle radius \(r=\dfrac{y}{2}=3-\dfrac{3x}{16}\).
Area \(A(x)=\dfrac{\pi r^{2}}{2}=\dfrac{\pi}{2}\Big(3-\dfrac{3x}{16}\Big)^{2}\).
Volume \(\displaystyle V=\int_{0}^{16}A(x)\,dx=\frac{\pi}{2}\int_{0}^{16}\!\Big(3-\frac{3x}{16}\Big)^{2}dx =\frac{\pi}{2}\Big[9x-\frac{9}{16}x^{2}+\frac{9}{768}x^{3}\Big]_{0}^{16}=24\pi\approx 75.398.\)
✅ Answer: (A)