Let $R$ be the surface in the first quadrant limited by the curves $x^2 + y^2 = 5, y = 2x$ and $x=0$. I'm asked to find the volume of the object that appears when $R$ is rotated about the y-axis.
Is there a smart way to represent the curve I'm interested in, in 1 function?
Since $R$ is rotated about the y-axis, I guess I have to make use of $$2\pi\int_{a}^{b} x (g(x)-f(x)) dx$$, but I'm not sure what my $g(x)$ and $f(x)$ would be in this case as I'm dealing with a circle in the first quadrant. Subtracting them like the formula suggest, would give me a function of two variables that I don't know how to integrate
The solution given by @MyGlasses is correct, but if you want to do the integration in $x$ then you have to realize that there is no $g(x)-f(x)$. $y(x)$ is a piecewise continuous function given by
$$y(x)=2xu(1-x)+\sqrt{5-x^2}u(x-1), \quad x\in[0,\sqrt{5}]$$
where $u(x)$ is Heaviside function.
Then, the volume is given by
$$ \begin{align} V &=2\pi\int_0^{\sqrt{5}} xy(x)~dx\\ &=2\pi\left[\int_0^1 2x^2~dx+\int_1^{\sqrt{5}} x\sqrt{5-x^2}~dx\right]\\ &=2\pi\left[\int_0^1 2x^2~dx+\frac{1}{2}\int_1^5 \sqrt{5-t}~dt\right],\quad t=x^2\\ &=\frac{20\pi}{3} \end{align} $$