the volume of revolving about the x-axis of the curve y=x^3, below the line y=1, and between x=0 and x=1?

Question

choices are pi/42, 0.143pi, pi/7, 0.643pi, 6pi/7

Answer 1

This should have been in the math section....In any case, the formula for the volume of a solid of revolution is:

∫π(f(x))^2dx [a, b]

Where a and b are the limits of integration. Since f(1) = (1)^3 = 1, we don't need to worry about working under the line y = 1, since the lines x = 1 and y = 1 at the point (1, 1) which is on the graph. So lets solve, shall we?

π∫(x^3)^2dx [0, 1]

π∫x^6dx [0, 1]

(π/7)x^7 [0, 1]

(π/7) * [1^7 - 0^7]

π/7

Your answer is π/7

Answer 2

The question is a little unclear. I assume you mean to revolve the area between the curves:

y = 1
y = x³
x = 0
x = 1

This area is above the curve y = x³ but below the line y = 1.

In that case use the washer method and integrate from x = 0 to x = 1.

Volume = ∫π(R² - r²)dx = ∫π[1² - (x³)²]dx

= ∫π[1 - x^6]dx = π[x - x^7/7] | [Evaluated from x = 0 to 1]

= π[1 - 1/7] - 0 = 6π/7