Hello,

I am stuck on Example 7 part a) in the link below. I have also attached my working out. Can anyone please help me out? Thanks

Two things:

\[ \text{Firstly, you've correctly noted that }A = \frac12 x^2 \tan \alpha.\\ \text{So your volume of a cross section should just be }\boxed{\delta V = \frac12 x^2 \tan \alpha\, \delta y}\\ \text{i.e. }\boxed{\delta V = \frac12 (r^2-y^2)\tan \alpha\, \delta y} \]

Not sure why you multiplied \(x^2\) to another \(r^2-y^2\) term in your expression. By doing so, because of the fact that \(x^2+y^2=r^2\), you're basically trying to deal with \(x^4\), or equivalently \((r^2-y^2)^2\)

\[ \text{Secondly, what is }a?\\ \text{The boundaries of integration (and also the sum) are just }-r\text{ and }r \]