▶️ Answer/Explanation
Solution
1. Region \( R \): Bounded by \( y = 4 \cos \left( \frac{\pi x}{4} \right) \) and \( y = (x – 2)^2 \), first quadrant.
2. Intersections: \( x = 0 \) and \( x = 2 \) (diagram shows two points).
3. Top curve: \( 4 \cos \left( \frac{\pi x}{4} \right) \) is above \( (x – 2)^2 \) from \( x = 0 \) to \( x = 2 \).
4. Triangle area: Isosceles right triangle, leg \( L = 4 \cos \left( \frac{\pi x}{4} \right) – (x – 2)^2 \), area \( \frac{1}{2} L^2 \).
5. Volume: \( \int_{0}^{2} \frac{1}{2} \left( 4 \cos \left( \frac{\pi x}{4} \right) – (x – 2)^2 \right)^2 \, dx \approx 1.775 \).
✅ A) 1.775.
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:
\(\frac{\pi }{4}\int_{0}^{6}x^{2}dx=\frac{\pi }{4}\left [ \frac{x^{3}}{3} \right ]^{6}_{0}=\frac{6^{3}\pi }{12}=18\pi \)