How do you compute the length of a curve $y^2=x^3 + x^2$

Question

So I came across this question:

consider the closed curve C defined by $y^2 = x^3 + x^2\ (-1\le x\le 0)$

i) compute the length of the curve

ii) find the area of the surface formed by revolving the curve completely about the $x$-axis

these steps are supposed to be done using Maple software, but unfortunately I couldn't figure them out :(

Any kind soul can help me out?

Answer 1

Conveniently, $y=0$ in precisely the case that $x=0$ or $x=-1$. Thus, the solid formed by revolving this curve is the same as the solid formed by revolving $y=\sqrt{x^3+x^2}$ (for which there is a formula in your cal 2 book [see ch8 here: http://ocw.mit.edu/resources/res-18-001-calculus-online-textbook-spring-2005/textbook/])

Also, the length of the curve will just be twice the length of $y=\sqrt{x^3+x^2}$, for which there is also a nice formula in your cal 2 book. (ch8 of the same book)

Cheers