Let T be the triangle 1\le x\le 2, 0\le y\le 3x-3. Find the volume that results by rotating the triangle T around the line x=-1.
To find the volume that results from rotating the triangle T around the line x = -1, we can use the method of cylindrical shells.
The key idea is to consider each vertical strip of the triangle T, parallel to the y-axis, and rotate it around the line x = -1 to form a cylindrical shell. We'll integrate the volumes of all these cylindrical shells to find the total volume.
First, let's determine the height of each cylindrical shell. The vertical strip of T, located at x = a (where 1 ≤ a ≤ 2), has a height given by the difference in y-values at the top and bottom of the strip. In this case, the bottom of the strip is defined by the equation y = 0, and the top is defined by y = 3x - 3. So the height of each cylindrical shell is (3a - 3) - 0 = 3a - 3.
Next, let's determine the radius of each cylindrical shell. The distance between the line x = a and the line x = -1 is |a - (-1)| = |a + 1|. Thus, the radius of each cylindrical shell is |a + 1|.
Finally, we integrate the volumes of all these cylindrical shells over the interval from 1 to 2, which corresponds to the range of x-values for the triangle T.
The volume V can be calculated as follows:
V = ∫[1, 2] π(radius)^2(height) dx
= ∫[1, 2] π(|x + 1|)^2(3x - 3) dx
Simplifying the expression, we get:
V = ∫[1, 2] π(x^2 + 2x + 1)(3x - 3) dx
= ∫[1, 2] (3πx^3 + 6πx^2 + 3πx - 3πx^2 - 6πx - 3π) dx
= ∫[1, 2] (3πx^3 - 3πx^2 - 3π) dx
Now, we can integrate the expression:
V = π[x^4 - x^3 - 3x] [1, 2]
V = π[(2^4 - 2^3 - 3(2)) - (1^4 - 1^3 - 3(1))]
V = π[(16 - 8 - 6) - (1 - 1 - 3)]
V = π(2)
V = 2π
Therefore, the volume that results from rotating the triangle T around the line x = -1 is 2π cubic units.