AP Calculus BC 8.8 Volumes with Cross Sections: Triangles and Semicircles - MCQs - Exam Style Questions
No-Calc Question
Let \(R\) be the region in the first quadrant bounded by the graph \(y=\sqrt{x-1}\), the \(x\)-axis, and the vertical line \(x=10\). Which of the following integrals gives the volume of the solid generated by revolving \(R\) about the \(y\)-axis?
(A) \(\displaystyle \pi\!\int_{1}^{10} (x-1)\,dx\)
(B) \(\displaystyle \pi\!\int_{1}^{10} \big(100-(x-1)\big)\,dx\)
(C) \(\displaystyle \pi\!\int_{0}^{3} \big(10-(y^{2}+1)\big)^{2}\,dy\)
(D) \(\displaystyle \pi\!\int_{0}^{3} \big(100-(y^{2}+1)^{2}\big)\,dy\)
(B) \(\displaystyle \pi\!\int_{1}^{10} \big(100-(x-1)\big)\,dx\)
(C) \(\displaystyle \pi\!\int_{0}^{3} \big(10-(y^{2}+1)\big)^{2}\,dy\)
(D) \(\displaystyle \pi\!\int_{0}^{3} \big(100-(y^{2}+1)^{2}\big)\,dy\)
▶️ Answer/Explanation
Detailed solution
Revolve around the \(y\)-axis ⇒ use washers (perpendicular slices).
Express \(x\) in terms of \(y\): from \(y=\sqrt{x-1}\) we have \(x=y^{2}+1\).
For \(y\in[0,\sqrt{10-1}]=[0,3]\):
• Outer radius \(R=10\) (to the line \(x=10\)).
• Inner radius \(r=y^{2}+1\) (to the curve \(x=y^{2}+1\)).
Volume: \[ V=\pi\int_{0}^{3}\!\big(R^{2}-r^{2}\big)\,dy =\pi\int_{0}^{3}\!\big(10^{2}-(y^{2}+1)^{2}\big)\,dy. \] ✅ Answer: (D)