Find the volume of the solid obtained by rotating the region in the first quadrant enclosed by the curves y= tanx and y=2cosx between the bounds 0 and pi/2 around y=-1
would it be
A= pi * ∫ 0 to 0.78 {(2cosx+1)^2 - (tanx+1)^2}
yes, if you include dx
The volume consists of washers of thickness dx and area pi(R^2-r^2) as you indicated above
To find the volume of the solid obtained by rotating the region in the first quadrant enclosed by the curves y=tanx and y=2cosx between the bounds 0 and pi/2 around the line y=-1, you can use the method of cylindrical shells.
First, let's sketch the region in the first quadrant between the curves y=tanx and y=2cosx:
To find where these two curves intersect, set them equal to each other:
tanx = 2cosx
Divide both sides by cosx:
sinx / cosx = 2
Use the trigonometric identity sinx/cosx = tanx:
tanx = 2
Now, solve for x by taking the arctangent of both sides:
x = arctan(2)
The bounds of integration for x are 0 and arctan(2).
Next, we need to set up the integral for finding the volume. The volume of the solid can be calculated using the formula:
V = ∫[a, b] 2π * radius * height dx
In this case, the radius is the distance from the rotation axis y=-1 to the curve y=tanx, and the height is the difference between the curves y=2cosx and y=tanx.
So, the integral becomes:
V = ∫[0, arctan(2)] 2π * (1 + tanx) * (2cosx - tanx) dx
Now, you can evaluate this integral to find the volume of the solid.