Consider the solid obtained by rotating the region bounded by the given curves about the line x = 1.
y = x, y = 0, x = 4, x = 6
Find the volume V of this solid
Volume= INT (Int radius of rotation*dArea)
= 2PI INT (x-1)(y)dx=2PI INT (x-1)x dx limits frm 1 to 6
Sorry for the other post.
To find the volume V of the solid, we can use the method of cylindrical shells.
The region bounded by the curves y = x, y = 0, x = 4, and x = 6 forms a rectangle in the xy-plane. When rotated about the line x = 1, this rectangle will form a solid with cylindrical shells.
First, let's find the height of each cylindrical shell. The height will be the difference between the y-values of the top and bottom curves at a given x-value.
The bottom curve is y = 0, so the height will be y = x - 0 = x.
The top curve is y = x, so the height will be y = x - 0 = x.
The height of each cylindrical shell is given by h = x.
Next, let's find the radius of each cylindrical shell. The radius will be the distance from the line x = 1 to the x-value on the rectangle.
Since the line x = 1 is the axis of rotation, the radius will be r = x - 1.
Now, let's express the volume of each cylindrical shell. The volume of a cylindrical shell is given by V = 2πrh, where r is the radius and h is the height.
Substituting r = x - 1 and h = x, we have V = 2π(x - 1)(x).
To find the total volume, we need to integrate this expression over the interval from x = 4 to x = 6, which represents the range of x-values in the region.
Thus, the volume V of the solid is given by:
V = ∫[4, 6] {2π(x - 1)(x) dx}
Evaluating this integral will give us the final answer for the volume of the solid.