Consider the region R in the first quadrant, which is located between the curves

Question

Consider the region 𝑅 in the first quadrant between the curves y = x.e^{(x^3)} and y=e for 0 ≤ 𝑥 ≤ 1.

a) Rotate R about the x-axis to find the volume of the solid.

b) Rotate R about the y-axis to find the volume of the solid.

Answer 1

It is always helpful to sketch a rough picture of the scenario. By some quick analysis, it can be shown that e ≥ xe^{x^3} ≥0 on [0,1].

1. Consider a slice of the region R rotated about the x-axis. The resulting figure is a washer, not a full disc. In this situation, the volume of the solid created is found by

V = π ∫_a^b [ f(x)² - g(x)² ] dx, where f is the greater function on the interval [a,b]

In our case,

V = π∫₀¹ (e² - (xe^{x^3})²) dx = π∫₀¹ e²dx - π∫₀¹ x²e^{2x^3}dx

I'll leave it to you to solve the integral. Hint: Use u-substitution on the second integral.

2) In this case, when our slice is rotated about the y-axis, when generate a shell. The formula in this case is

v = 2π ∫_a^b x[ f(x)-g(x)] dx, again where f is the greater function. so

V = 2π ∫₀¹ x[e - xe^{x^3}] dx = 2π ∫₀¹ xe dx - 2π ∫₀¹ x²e^{x^3}dx

As before, compute the definite integral and you'll find your volume.