Find the volume of the solid obtained by rotating the region bounded by y=6x^2, x=1,and y=0, about the x-axis.

Answer 1

y = 6x^2, x=1, and y=0

Essentially, you're given the three pieces of information you need to solve for this problem, since you know the equation of the function y, which can also be written as f(x), as well as the x-values for your boundaries. y = 0 represents the x-axis, and x = 1 is the vertical line along that value, so you know that the definite integral will be going from 0 to 1.

From here you can just use the formula for volume to obtain your answer:

V = pi x Integral as x goes from 0 to 1 (f(x))^2 dx
V = pi x Integral (6x^2)^2 dx
V = pi x Integral 36x^4 dx

= pi x 36 x x^5/5 = 36pi(x^5/5) [as x goes from 0 to 1]

Now you can just use the definite integral and plug in the values, while leaving the constants 36 and pi outside until the very end.

36pi[(1^5/5) - (0^5/5)]
36pi(1/5 - 0)
36pi/5

Your answer should be 36pi/5. It should be alright to leave it in the form instead of multiplying out pi, but either format should work.

Hope that helps!

Answer 2

4

Answer 3

Use the disc method.

Integrate from x = 0 to x = 1.

∫πy²dx = ∫π(6x²)²dx = 36π∫x^4 dx

= 36πx^5/5 | [Evaluated from 0 to 1]

= 36π/5

Answer 4

36pi/5.