>Find the volume of the solid generated by revolving about the y-axis the region bounded by the curve $y=x^3$ and the lines $y=0$ and $x=2$

Question

Find the volume of the solid generated by revolving about the y-axis the region bounded by the curve $y=x^3$ and the lines $y=0$ and $x=2$

My attempt: Take radius $r=(2-y^{1/3})$

Then volume $V=\int_{0}^{8}r^2dy=16\pi/5$

I saw the answer to this question online and it is $64\pi/5$

The method went like this :$V=\int_{0}^{8}2dy- \int_{0}^{8}y^{2/3}dy=64\pi/5$

However I can't figure out why my method if wrong? What volume is my answer giving?

Answer 1

What you have actually computed is the volume of the region obtained by rotating around the $y$-axis the region bounded by the curve $y=(2-x)^3\left(\iff x=2-\sqrt[3]y\right)$, and by the lines $x=2$ and $y=0$ (see the picture below). It is natural that it leads to a different answer.

Answer 2

Try defining a function $g(x)=f^{-1}(x)$. Then the volume of the curve about the y-axis is $V = \pi\cdot \int_{0}^{8}2^{2}-g(x)^{2}dx$ In your calculation you do not separate the two integrals, there you write $r = (2-g(x))$ as one function as Carlos Santos pointed out above.

In general, we know from Calc that the area between two curves is defined as $\int f(x)-g(x) dx \Longleftrightarrow \int f(x) - \int g(x)$. Similarly for the Volume of the cross-section between two curves $f(x)$ and $g(x)$ is $$\pi \cdot \int f(x)^2 -g(x)^2dx$$