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 isosceles right triangles with one leg on the base of the solid.
A) $\frac { 9 } { 4 } \pi$
B) $\frac { 3 } { 2 } \pi$
C) $\frac { 9 } { 8 } \pi$
D) $2 \pi$

Question

Quizplus · Accepted Answer

The area of each isosceles right triangle cross-section is $\frac{1}{2} \times (3 \cos x)^2$. Integrating this from $0$ to $\pi/2$ gives the volume: $\int_0^{\pi/2} \frac{1}{2} (3 \cos x)^2 \, dx = \frac{9}{2} \int_0^{\pi/2} \cos^2 x \, dx$. Using the identity $\cos^2 x = \frac{1 + \cos(2x)}{2}$, the integral becomes $\frac{9}{4} \left[ x + \frac{1}{2} \sin(2x) \right]_0^{\pi/2} = \frac{9}{8} \pi$.

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

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