AP Calculus AB : 8.8 Volumes with Cross Sections: Triangles and  Semicircles- Exam Style questions with Answer- MCQ

Question

Let R be the region in the first quadrant bounded by the graphs of \(y=4\cos (\frac{\pi x}{4})\) and \(y=(x-2)^{2}\) , as shown in the figure above. The region R is the base of a solid. For the solid, each cross section perpendicular to the x-axis is an isosceles right triangle with a leg in region R. What is the volume of the solid?
A 1.775
B 3.549
C 4.800
D 5.575

▶️Answer/Explanation

Ans:A

Question

 Consider a solid whose base is the region bounded by the curve \(\log _{2}(x)\)  and the lines y = -1 and x = 2. If parallel cross sections perpendicular to the y-axis are regular triangles, what is the volume of
the solid?
(A) 1.212
(B) 0.658
(C) 0.887
(D) 0.713

▶️Answer/Explanation

Ans:(C)

Question

 If a solid S has as its base a triangular region with vertices (0, 0), (3, 0), and (0, 1), and its parallel cross sections perpendicular to the x-axis are squares, what is the volume of S ?
(A) \(\frac{72}{5}\)
(B) 7
(C) \(\frac{45}{11}\)
(D) 5

▶️Answer/Explanation

Ans:(D)

Question

 Find the volume of the solid obtained by rotating the region bounded by a right triangle with vertices at the origin, (6, 0), and (6, 3) about the x-axis.
(A) \(9\pi \)
(B) \(18\pi \)
(C) \(27\pi \)
(D) \(36\pi \)

▶️Answer/Explanation

Ans:(B)

See the figure below.

A cone generated by rotating \(y=\frac{x}{2}\) about the x-axis.
An easy algebraic solution follows from the observation that the revolution of the triangle fills out a cone of height 6 and base radius 3. Hence,
\(V_{Cone}=\frac{1}{3}\pi .3^{2}.6=18\pi \)
For a solution using calculus, note that the hypotenuse of the triangle is described by the equation \(y=\frac{1}{2}x\) from x = 0 to x = 6. Therefore, the volume of the cone is obtained by the following computation:
\(\frac{\pi }{4}\int_{0}^{6}x^{2}dx=\frac{\pi }{4}\left [ \frac{x^{3}}{3} \right ]^{6}_{0}=\frac{6^{3}\pi }{12}=18\pi \)

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