How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region #y=x^(1/2)#, #y=0#, and #x=4# rotated about the x axis?

Question

Ezra Adams · Accepted Answer

This solid to-be-revolved looks like: graph{(y - sqrtx)(y)sqrt(4-(x-0))/sqrt(4-(x-0)) <= 0 [-5, 5, 0, 5]}  If you want to do it with the shell method, convert your functions to their inverses. You get:
#y = x^2#
#x = 0#
#y = 4# Now your domain is your range and your interval is #[0,2]#. graph{(y - x^2)sqrt(4-(y-0))/sqrt(4-(y-0)) >= 0 [0, 8, -1, 5]}  #V = 2piintxf(x)dx# #f(x) = x^2 in [0,2]#, bounded from above by #y = 4#, thus #f(x) = 4-x^2# #V = 2piint_0^2 x*(4-x^2)dx# #= 2pi[2x^2 - x^4/4]|_(0)^(2)# #= 2pi[(2(2)^2 - (2)^4/4) - cancel((2(0)^2 - (0)^4/4))]# #= color(blue)(8pi "u") ~~ 25.1327 "u"# Note: if you evaluated this using #f(x) = x^2# instead, you would coincidentally get the same answer, but for the wrong work, so if you did that on a test, you wouldn't get full credit.

Luke Alvarez · Answer

undefined To use the shell method to find the volume of the solid generated by revolving the given plane region about the x-axis, follow these steps:

1. Identify the interval of integration for x. In this case, the interval is from x = 0 to x = 4, as given by the intersection points of y = x^(1/2) and the x-axis.

2. Determine the radius of each shell. Since we are revolving about the x-axis, the radius of each shell is the distance from the axis of rotation (x-axis) to the curve. Therefore, the radius is simply x.

3. Determine the height of each shell. The height of each shell is the difference between the upper and lower curves at each x-value. In this case, the upper curve is y = x^(1/2) and the lower curve is y = 0. So, the height is simply y = x^(1/2).

4. Setup the integral to compute the volume using the shell method:

$$V = \int_{0}^{4} 2\pi x (x^{1/2}) \, dx$$

5. Evaluate the integral:

$$V = 2\pi \int_{0}^{4} x^{3/2} \, dx$$

$$V = 2\pi \left[\frac{2}{5}x^{5/2}\right]_{0}^{4}$$

$$V = 2\pi \left(\frac{2}{5} \cdot 4^{5/2} - \frac{2}{5} \cdot 0^{5/2}\right)$$

$$V = 2\pi \left(\frac{2}{5} \cdot 32 - 0\right)$$

$$V = \frac{64}{5}\pi$$