Find the volume of the solid generated by the region in the first quadrant bounded above by the 3x+y=6, below by the x-axis, and on the left by the y-axis, about the line x= -2.
If we include the area out to x = -2, we have
v = ∫[0,6] πr^2 dy
where r = 2+x = 2+(6-y)/3 = 4-y/3
v = π∫[0,6](4-y/3)^2 dy = 56π
But we have to subtract out the interior cylinder of radius 2 and height 6, or 24π, leaving us with just 32π generated by the rotating triangle.
To find the volume of the solid generated by rotating a region in the first quadrant bounded by curves about a vertical or horizontal line, we can use the method of cylindrical shells.
Let's consider the given region bounded above by the equation 3x + y = 6 and below by the x-axis. Also, it is bounded on the left by the y-axis.
First, we need to sketch the region. By rearranging the equation 3x + y = 6, we can rewrite it as y = 6 - 3x. This is a straight line with a y-intercept of 6 and a slope of -3. It intersects the x-axis at x = 2. Therefore, the region of interest lies between the x-axis and the line y = 6 - 3x in the first quadrant.
Next, we want to rotate this region about the line x = -2. Since x = -2 is a vertical line, the solid generated will have a cylindrical shape.
To use the method of cylindrical shells, we break the solid into infinitely thin cylindrical shells with height Δx and radius r.
The radius r of each shell is the distance from the vertical line x = -2 to the corresponding x-coordinate on the line y = 6 - 3x. Using the formula for the distance between two lines parallel to the x-axis, r = x + 2.
Since the region is bounded by the x-axis and the line y = 6 - 3x, the height of each shell, Δx, will range from 0 to the x-coordinate on y = 6 - 3x.
To find the volume of each shell, we multiply its circumference (2πr) by its height Δx.
The volume of each shell is given by dV = 2πr * Δx.
To find the volume of the entire solid, we integrate the volume of each shell. The integration is done with respect to x, from 0 to 2 (the x-coordinate where the region intersects the x-axis):
V = ∫(0 to 2) 2πr * Δx
Substituting r = x + 2 and Δx = 6 - 3x, the equation becomes:
V = ∫(0 to 2) 2π(x + 2)(6 - 3x) dx
Evaluating this integral will give you the volume of the solid generated by the region in the first quadrant, bounded above by the line 3x + y = 6, below by the x-axis, and on the left by the y-axis, about the line x = -2.