What is the volume of the solid produced by revolving #f(x)=xe^x-(x/2)e^x, x in [2,7] #around the x-axis?

Question

Ezekiel Alexander · Accepted Answer

The volume of the solid produced by revolving around the x-axis the trapezoid of the function #f(x)# is calculated using the formula:

#V = pi int_a^b f^2(x)dx# #V= pi int_2^7 (xe^x-(x/2)e^x)^2dx# #V= pi int_2^7 (x^2e^(2x)-2xe^x(x/2)e^x+x^2/4e^(2x))dx# #V= pi int_2^7 (x^2e^(2x)-x^2e^(2x)+x^2/4e^(2x))dx = pi int_2^7 x^2/4e^(2x)dx # Substitute #2x=t#: #V = pi/32 int_1^(7/2) t^2e^tdt # The integral is solved iteratively by parts: #int t^2e^tdt = int t^2d(e^t) =t^2e^t - int 2te^tdt = t^2e^t - 2te^t +2inte^tdt = e^t(t^2-2t+2)# Finally: #V = pi/32[e^(7/2)(49/4-7+2)-e(1-2+2)] = pi/32[e^(7/2)(49-28+8)/4-e] = (pi*e)/32 (29/4e^(5/2)-1) #

Axel Alvarez · Answer

undefined To find the volume of the solid produced by revolving the function $ f(x) = xe^x - \frac{x}{2}e^x $ on the interval $ x \in [2,7] $ around the x-axis, you can use the method of cylindrical shells:

1. Determine the radius of each cylindrical shell. Since we're revolving around the x-axis, the radius is the distance from the axis to the function, which is $ f(x) $.

2. Determine the height of each cylindrical shell. The height of each shell is the differential element along the x-axis, which is $ dx $.

3. Write the volume element $ dV $ for each cylindrical shell as $ dV = 2\pi x f(x) dx $, as the circumference of a cylindrical shell is $ 2\pi x $ and the height is $ f(x) $.

4. Integrate $ dV $ from $ x = 2 $ to $ x = 7 $ to find the total volume:
   $$ V = \int_{2}^{7} 2\pi x f(x) \, dx $$

5. Compute the definite integral to find the volume.

6. The resulting value will be the volume of the solid produced by revolving $ f(x) $ around the x-axis on the given interval.