Question

Find the volume of the solid obtained by rotating the region bounded by y = x and y = x^3 about the line y = 3, where x>or equal to 0

Answer 1

Step 1: Rearrange the equation in respect to (y) since we are rotating via the y-axis. Thus, x=y and x=y^(1/3)

\[(\pi) \int\limits_{0}^{3} ((y)-(y ^{(1/3)})^2 dy\]

After integrating and solving the answer should be:

\[(\pi) ((9*3^{2/3}/5)-(54*3^{1/3}/7)+9)\]

Answer 2

If you look at the graph of y=x and y=x^3 you will see that rotating on y=3 forms a cone. Which lets us use the disk method of finding area by slicing. Disk Method = \[\int\limits_{a}^{b} (\pi) (f(x))^2 dx\]

Answer 3

whats the final answer though?

Answer 4

I approached this with the washer method. I'm rsuty so I don;t have certainty in my answer:
\[\pi \int\limits_{0}^{1} [(3-x^3)^2-(3-x)^2]dx\]
\[\pi \int\limits_{0}^{1}[(9-6x^3+x^6)-(9-6x+x^2)]dx\]
\[\pi \int\limits_{0}^{1} (-6x^3+x^6+6x-x^2)dx\]
\[\pi [-\frac{3}{2}x^4+\frac{1}{7}x^7+3x^2-\frac{1}{3}x^3]_{0}^1\]
\[\pi (-\frac{3}{2}+\frac{1}{7}+3-\frac{1}{3})=\frac{55}{42}\pi\]