Let $R$ be the region bounded by the following curves. Find the volume of the solid generated when $R$ is revolved about the $x$ -axis. Recall that $\cos^{2} x = \frac{1}{2} (1 + \cos 2 x)$ .
$y = \cos 2 x, y = 0, x = 0$
The volume of the region revolved about the $x$ -axis is cubic units. (Type an exact answer.)

Answer

Expert–verified

Hide Steps

Answer

$V = \frac{5 π}{16}$ cubic units.

Steps

Step 1 ：Given the curve $y = \cos 2 x$ , the x-axis, and the y-axis, we need to find the volume of the solid generated by revolving the given region around the x-axis.

Step 2 ：Using the disk method, the area of a cross-sectional disk is given by $A (x) = π y^{2}$ . Substituting the given equation for y, we get $A (x) = π (\cos 2 x)^{2}$ .

Step 3 ：We know that $\cos^{2} x = \frac{1}{2} (1 + \cos 2 x)$ , so we can simplify the area function to $A (x) = π \frac{1}{2} (1 + \cos 4 x)$ .

Step 4 ：Integrate the area function along the x-axis from $x = 0$ to $x = \frac{π}{4}$ to find the volume: $V = \int_{0}^{\frac{π}{4}} A (x) d x$ .

Step 5 ：Simplify the integral to get the volume: $V = \frac{π}{4} + \frac{π}{16}$ .

Step 6 ： $V = \frac{5 π}{16}$ cubic units.

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the x-axis. Recall that cos2⁡x=12(1+cos⁡2x).y=cos⁡2x,y=0,x=0The volume of the region revolved about the x-axis is cubic units. (Type an exact answer.)

Answer

Popular Problems