Use washer and shell method to find $k$

Question

Let $R$ denote the first quadrant region bounded by $y=k-kx^2$, and $y=kx^2$. $k> 0, k \in \mathbb{R}$. If $V_1$ is the volume obtained by rotating $R$ about the line $y=0$, and $V_2$ is the volume obtained by rotating R about the line $x=-2$, and $V_1=V_2$, then what is $k$?

My plan is to evaluate exact volume of $V_1$ and $V_2$, then make the equation $V_1=V_2$ and solve for $k$. The first one $V_1$ can use the washer method, and second one might be shell, since it is rotating around a vertical line $x = 2$.

Answer 1

Note that the two curves meet at $x= \frac1{\sqrt2}$. Then, washer-integrate $V_1$ and shell $V_2$ over $x\in(0,\frac1{\sqrt2})$

\begin{align} &V_2-V_1\\ =&2\pi\int_0^{1/\sqrt2} (x+2)(k(1-x^2)-kx^2))dx - \pi\int_0^{1/\sqrt2}(k^2(1-x^2)^2-k^2 x^4)dx\\ = &\left(\frac{4\sqrt2}3\pi k +\frac\pi4k \right)- \left(\frac{\sqrt2}3\pi k^2 \right)= 0 \end{align} which yields

$$k= 4+ \frac{3\sqrt2}{8}$$

Answer 2

The region being revolved is bound by curves $y = k - kx^2$ and $y = kx^2$ for $k \gt 0$.

At intersection of both curves, $k - kx^2 = kx^2 \implies x = \frac{1}{\sqrt2}$ as we are in first quadrant.

For revolution around $y = 0$, we apply washer method. Radius of revolution $y$ is between $kx^2 \leq y \leq k-kx^2$.

$V_1 = \displaystyle \int_{0}^{1/\sqrt2}\int_{kx^2}^{k-kx^2} 2 \pi \ y \ dy \ dx = \frac{\sqrt2}{3} \pi \ k^2$

For revolution around $x=-2$, we apply shell method. The size of shell is given by $k-kx^2 - kx^2 = k - 2kx^2$. Radius of revolution is $2+x$.

$V_2 = \displaystyle \int_{0}^{1/\sqrt2} 2 \pi \ (2+x) (k-2kx^2) \ dx = \frac{3 + 16\sqrt2}{12} \pi \ k$

When $V_1 = V_2, k = \frac{3 + 16\sqrt2}{4\sqrt2} \approx 4.53$.