- #1

karush

Gold Member

MHB

- 3,269

- 5

about y-axis.Given the boundares of }\\$

\begin{align*} \displaystyle

y&=4x-x^2\\

y&=x\\

\end{align*}

Ok, I presume since this is rotated around the y-axis that we have to rewrite the equations in terms of y

$y=-(x^2-4x)$

$4-y=(x-2)^2$

$\sqrt{4-y}=x-2$

$\sqrt{4-y}+2=x$

and

$x=y$

But the graph is not complete and is this not the same volume if it were rotated around the x-axis.

I thought this was going to be very easy, but going around the y-axis was confusing.