Problem
Suppose that $Y_1,Y_2,Y_3$ denote a random (independent) sample of size $3$ from a distribution with parameter $\lambda$ defined by the following probability density function:
$$\begin{equation} f(y)=\begin{cases} \lambda y^{\lambda−1} , & \text{$0 < y < 1$, $\lambda>0$}\\ 0, & \text{elsewhere} \end{cases}. \end{equation}$$
The observed values of the sample are $y_1 = 0.1, y_2 = 0.2 \ \mathrm{and} \ y_3 = 0.3.$
$(a)\quad$ Let $W_i = − \ln(Y_i)$. Find the distribution of each $W_i$.
$(b)\quad$ Derive the distribution of $U = 2\lambda(W_1 + W_2 + W_3).$
$(c)\quad$ Using $(b)$, or otherwise, find an exact central $95$% confidence interval for $\lambda$ and evaluate it for the sample.
My answers
(a)
$$\begin{aligned} y_1 & = e^{-w_1} \\ \frac{dy_1}{dw_1} & = -e^{-w_1} \end{aligned}$$ As $y_1$ is a decreasing function of $w_1$, so by using the transformation method, we have $$\begin{aligned} f_{W_1}(w_1) & = f_Y(y_1) \left| \frac{dy_1}{dw_1}\right| \\ & = \lambda e^{-w_1 \lambda} ,\ \mathrm{where} \ 0 < e^{-w_1} < 1 \end{aligned}$$
and $$W_1 \sim exp(\lambda), \lambda > 0.$$ By symmetry, it follows that $$W_i \sim exp(\lambda), \lambda > 0.$$
(b)
Let $$J = \sum_{i=1}^{3} W_i.$$ Since $$W_i \overset{iid}{\sim} Exp (\lambda),$$ so $$\begin{aligned} J & \sim Gamma \left( 3, \frac{1}{\lambda} \right)\\ \implies U & = 2 \lambda J\\ \implies U & \sim Gamma(3,2) \end{aligned}$$
As I have just covered these topics, I am uncertain of my calculations and approaches for $(a)$ and $(b)$. Am I on the right track? If I am, then how should I continue for $(c)$ as it seems my answer to $(b)$ does not contain $\lambda$ anymore? Any intuitive explanations will be highly appreciated!
Edit
Following the comments, I have tried the following but am unsure whether it is correct.
$(c)$
$U = 2 \lambda J \sim Gamma (3,2)$
$$0.95 = \mathbb{P} \left( \frac{0.025}{2J} \leq \lambda \leq \frac{0.975}{2J} \right)$$
Thus, it follows that exact central $95 %$ confidence interval for $\lambda$ is $$\left[\frac{0.025}{2(W_1 + W_2 + W_3)} ,\frac{0.975}{2(W_1 + W_2 + W_3)} \right]$$ and the exact central $95%$ confidence interval for $\lambda$ for the sample is $$\left[\frac{0.025}{-2 \ln (0.006)} ,\frac{0.975}{-2 \ln(0.006)} \right].$$