Find the volume of the described solid.
-The base of a solid is the region between the curve $y = 3 \cos x$ and the $x$-axis from $x = 0$ to $x = \pi / 2$. The cross sections perpendicular to the $x$-axis are squares with diagonals running from the $x$-axis to the curve.
A) $\frac { 9 } { 4 } \pi$
B) $\frac { 9 } { 8 } \pi$
C) $\frac { 3 } { 2 } \pi$
D) $2 \pi$

Question

Quizplus · Accepted Answer

The volume of the solid can be calculated by integrating the area of the cross-sections along the given interval. The cross-sections are squares, and the length of a diagonal of each square is given by the distance from the $x$-axis to the curve, which is $y = 3 \cos x$. The length of a side of each square, $s$, is related to the diagonal, $d$, by the formula $d = s\sqrt{2}$, because the diagonal of a square splits it into two 45-45-90 right triangles. Therefore, $s = \frac{d}{\sqrt{2}} = \frac{3\cos x}{\sqrt{2}}$. The area of each square is $s^2 = \left(\frac{3\cos x}{\sqrt{2}}\right)^2 = \frac{9\cos^2 x}{2}$. To find the volume, integrate this area from $x = 0$ to $x = \frac{\pi}{2}$:$$V = \int_{0}^{\frac{\pi}{2}} \frac{9\cos^2 x}{2} dx$$Using the double angle identity $\cos^2 x = \frac{1 + \cos 2x}{2}$, the integral becomes:$$V = \frac{9}{2} \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos 2x}{2} dx = \frac{9}{4} \left[ x + \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}}$$Evaluating this from $0$ to $\frac{\pi}{2}$ gives:$$V = \frac{9}{4} \left[ \frac{\pi}{2} + 0 - 0 \right] = \frac{9}{4} \cdot \frac{\pi}{2} = \frac{9\pi}{8}$$Therefore, the volume of the solid is $\frac{9}{8}\pi$, which corresponds to choice B.

Find the Volume of the Described Solid y=3cos⁡xy = 3 \cos xy=3cosx

Find the Volume of the Described Solid $y = 3 \cos x$