Good day.
I am trying to solve this question:
"Find the volume of the solid obtained by revolving the indicated region about the given line. The region is bounded by the curves $y=x\sqrt{2-x}$ , $y=6-x$ , and x-axis. It is rotated in the y-axis."
I assigned the $f(x)$ to the $y=x\sqrt{2-x}$ and the $g(x)$ to the $y=6-x$
I learned that when it is rotated in the y-axis, I need to change a function that is terms of x to a function that is in terms with y.
I may hard to find when I tried to change the function, $y=x\sqrt{2-x}$ , to a f(y). While I solved the $g(x)$ from $y=6-x$ to $x=6-y$.
I tried to solve the $y=x\sqrt{2-x}$ first. The first step that I did is squaring the both sides to cancel the square root. And the result is $y^2=2x^2-x^3$. What steps do I need to do to cancel the powers of the x for it to become $f(y)$ ?
Thank you for answering.
From $y=x\sqrt{2-x},$ you get $y^2=x^2(2-x).$ Rearranging terms, $x^3-2x^2+y^2=0.$ Substituting $x=u+\frac{2}{3}$, $$ 0=(u+\frac{2}{3})^3-2(u+\frac{2}{3})^2+y^2=u^3-\frac{4}{3}u+(y^2-\frac{16}{27}). $$
Then from the formula at https://en.wikipedia.org/wiki/Cubic_equation#Cardano's_formula for solving "depressed cubics", $$ u =\left(-\frac{y^2-\frac{16}{27}}{2} +\sqrt{\frac{(y^2-\frac{16}{27})^2}{4}-\frac{64}{729}}\right)^{\frac{1}{3}} +\left(-\frac{y^2-\frac{16}{27}}{2}-\sqrt{\frac{(y^2-\frac{16}{27})^2}{4}-\frac{64}{729}}\right)^{\frac{1}{3}}. $$ Then $$ x=u+\frac{2}{3}. $$ The other two solutions can be obtained from multiplying each of the two cube roots by conjugate third roots of unity. Some of these values will be solve $y=x\sqrt{2-x},$ some will solve $y=-x\sqrt{2-x}.$