Find the volume of the described solid.
-The base of a solid is the region between the curve $y = 6 \cos x$ and the $x$-axis from $x = 0$ to $x = \pi / 2$. The cross sections perpendicular to the $x$-axis are squares with bases running from the $x$-axis to the curve.
A) $3 \pi$
B) $18 \pi$
C) $9 \pi$
D) $\frac { 35 } { 4 } \pi$

Question

Quizplus · Accepted Answer

The volume of the solid can be calculated by integrating the area of the cross-sections along the given interval. Since the cross-sections are squares with sides equal to $y = 6 \cos x$, the area of each square is $A(x) = (6 \cos x)^2 = 36 \cos^2 x$. The volume $V$ is then given by the integral of $A(x)$ from $x = 0$ to $x = \pi / 2$, which is $V = \int_{0}^{\pi/2} 36 \cos^2 x \, dx$. Using the double angle formula, $\cos^2 x = \frac{1 + \cos(2x)}{2}$, the integral becomes $V = 36 \int_{0}^{\pi/2} \frac{1 + \cos(2x)}{2} \, dx$, which evaluates to $9 \pi$.

Find the Volume of the Described Solid y=6cos⁡xy = 6 \cos xy=6cosx

Find the Volume of the Described Solid $y = 6 \cos x$