How do you find the volume of the solid generated by revolving the region bounded by the curves #y = x-2#, the #x#-axis, #x=2#, and #x=4# rotated about the #x=-1#?

Question

Elijah Abram · Accepted Answer

Let's see what this looks like: graph{(y+2-x)(sqrt(2-(x-2)))/(sqrt(2-(x-2))) <= 0 [-5, 6, -0.3, 6]}  Revolving around #x = -1# means doing the shell method makes this easier (you're revolving to make a shell, and you don't have to convert to #f(y)# functions). #V = 2piint_a^b xf(x)dx# where:
#x# is the radius of the shell
#f(x)# is the thickness of the shell
#2pi# indicates that the shell is fully revolved (#2pi# radians = 1 full revolution) Now it's just a matter of figuring out what your equations are. #f(x) = x-2#
(the thickness is from the inside surface of the shell to the outside surface, which is what #f(x)# from 2 to 4 revolves to become) Interval: #[2,4]#
(implied by #x = 2# and #x = 4# in the question) The radius of the shell is determined by the distance from the axis of #x = -1# to the current value of #x# in the interval #[2,4]#. Therefore, the radius is #x - (-1) = x+1#. Overall, we get: #V = 2piint_2^4 (x+1)(x-2)dx# #= 2piint_2^4 x^2 - x - 2dx# #= 2pi[x^3/3 - x^2/2 - 2x]|_(2)^(4)# #= 2pi[(4^3/3 - 4^2/2 - 2(4))-(2^3/3 - 2^2/2 - 2(2))]# #= 2pi[(64/3 - 16/2 - 8)-(8/3 - 4/2 - 4)]# #= 2pi[(64/3 - 48/3)-(8/3 - 18/3)]# #= 2pi[64/3 - 48/3 - 8/3 + 18/3]# #= 2pi[26/3]# #color(blue)(= (52pi)/3 "u"^3)#

Ezra Adams · Answer

undefined To find the volume of the solid generated by revolving the region bounded by the curves y = x - 2, the x-axis, x = 2, and x = 4 rotated about the line x = -1, you would use the method of cylindrical shells.

The formula for the volume using cylindrical shells is:

V = ∫[a, b] 2π * radius * height * dx

In this case, the region bounded by the curves is between x = 2 and x = 4. The radius of each shell is the distance from the axis of rotation (x = -1) to the curve, which is x + 1. The height of each shell is the y-value of the curve, which is (x - 2).

Therefore, the integral to find the volume would be:

V = ∫[2, 4] 2π * (x + 1) * (x - 2) dx

You would then integrate this expression over the interval [2, 4] to find the volume of the solid.

William Abdalla · Answer

undefined To find the volume of the solid generated by revolving the region bounded by the curves $ y = x - 2 $, the x-axis, $ x = 2 $, and $ x = 4 $ rotated about the line $ x = -1 $, you can use the method of cylindrical shells.

1. First, determine the limits of integration. The region is bounded by $ x = 2 $ and $ x = 4 $, so these will be the limits of integration.

2. Next, determine the radius of each shell. Since we are rotating about the line $ x = -1 $, the distance from each shell to the axis of rotation is $ x - (-1) = x + 1 $.

3. Determine the height of each shell. The height of each shell is the difference in the y-values of the upper and lower curves. The upper curve is $ y = x - 2 $ and the lower curve is the x-axis, so the height of each shell is $ (x - 2) - 0 = x - 2 $.

4. Set up the integral for the volume using the formula for the volume of a cylindrical shell:
$$ V = \int_{a}^{b} 2\pi \cdot (\text{radius}) \cdot (\text{height}) \, dx $$

5. Substituting the expressions for radius and height:
$$ V = \int_{2}^{4} 2\pi \cdot (x + 1) \cdot (x - 2) \, dx $$

6. Integrate the expression to find the volume:
$$ V = \int_{2}^{4} 2\pi \cdot (x^2 - x - 2) \, dx $$

$$ V = 2\pi \cdot \left[ \frac{x^3}{3} - \frac{x^2}{2} - 2x \right]_{2}^{4} $$

$$ V = 2\pi \left( \frac{64}{3} - 8 - 8 - \frac{8}{3} + 4 + 4 \right) $$

$$ V = 2\pi \left( \frac{64}{3} - \frac{8}{3} - 16 \right) $$

$$ V = 2\pi \left( \frac{56}{3} - 16 \right) $$

$$ V = 2\pi \left( \frac{56}{3} - \frac{48}{3} \right) $$

$$ V = 2\pi \cdot \frac{8}{3} $$

$$ \boxed{V = \frac{16\pi}{3}} $$