How to apply the volume of solid revolution formula in this case?

Question

I'm trying to find the volume of a object using the solid revolution formula: $$\int_a^b \frac{\pi}{2}y^2 \;\mathrm{dx}$$ The shape of my object can be given by this equation: $$(x^2+y^2)^2=4x^3-2xy^2$$ Where the domain is (0, 1). However, the function's output exist both above and below the x-axis, a situation I haven't encountered in my class. I tried to isolate y, but couldn't find a way to do so. Is there a way to write the equation as a function of y? If no, can I still find a way to calculate the volume created by rotating the positive (or negative) half of the function around the x-axis?

Answer 1

Given the formula, $\displaystyle{\int_a^b \frac{\pi}{2}y^2 \;\mathrm{dx}}$ , is employed, it is being suggested you want find the volume of revolution around the $x$-axis. Hence, I assume your domain is w.r.t to $x$; $x \in (0, 1)$.

Your equation is quadratic in $y^2$;

$$\left(x^2 + y^2\right)^2 = 4x^3 - 2xy^2 \longrightarrow x^4 - 4x^3 + y^2 \cdot \left(2x^2 + 2x\right) + \left(y^2\right)^2 = 0$$

Where the quadratic formula enables us to express $y^2$ as function of $x$:

$$y^2 = \dfrac{-\left(2x^2+2x\right) \pm \sqrt{\left(2x^2+2x\right)^2-4\cdot 1 \cdot \left( x^4 - 4x^3\right)}}{2} = -x(x + 1) \pm x\sqrt{6x+1}$$

A nifty trick to understand here is, you do not need to waste effort in solving "the equation as function of $y$", as your integrand is $y^2$. This reduces effort in dealing with $2$ solutions instead of $4$. Out of the 2 solutions, $y^2 = -x(x + 1) - x\sqrt{6x+1} < 0, x > 0$, which would result in a negative volume.

Hence, the correct integrand would be $y^2 = x\sqrt{6x+1}-x(x+1)$. Can you continue the remaining integration?