207k views
5 votes
This is the question

This is the question-example-1
User Kmas
by
6.7k points

1 Answer

7 votes

Check the picture below, so the lines look more or less like so, so the shape that we'll be getting will be the shape of a bowl with a hole in it, thus we'll use the washer method.

the way I use to get the larger Radius is simply using the "area under the curve" method, namely f(x) - g(x), where g(x) in this case is the axis of rotation.

so to get "R" and "r" we can get it by


\stackrel{y = 1}{(1)}~~ - ~~\stackrel{\stackrel{\textit{axis of rotation}}{y = -3}}{(-3)}\implies 1+3\implies \stackrel{R}{4} \\\\\\ \stackrel{y = \ln(x)}{\ln(x)}~~ - ~~\stackrel{\stackrel{\textit{axis of rotation}}{y = -3}}{(-3)}\implies \stackrel{r}{ln(x)+3} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{R^2}{(4)^2}~~ - ~~\stackrel{r^2}{[\ln(x)+3]^2}\implies 16~~ - ~~[\ln^2(x)+6\ln(x)+9] \\\\\\ \ln^2(x)+6\ln(x)+7~\hfill \boxed{\displaystyle\int~[\ln^2(x)+6\ln(x)+7]dx}

This is the question-example-1
User Tborzecki
by
6.0k points