Calculate power of the test $H_0$: $\sigma^2 \leq \sigma_0^2$ vs. $H_1$: $\sigma^2 > \sigma_0^2$ for $\mathcal{N}(0, \sigma^2)$ data. Let $X_1, \dots, X_n \overset{\text{iid}}{\sim} \mathcal{N}(0, \sigma^2)$. We consider the testing problem $H_0$: $\sigma^2 \leq \sigma_0^2$ vs. $H_1$: $\sigma^2 > \sigma_0^2$ and the statistical test $\delta:\mathbb{R}^n \rightarrow \{0, 1\}$ defined as follows:
$$\delta(x) = \begin{cases}
1 &\mbox{if } \ \frac{1}{\sigma_0^2}\sum_{i=1}^n x_i^2 \geq \chi^{2-}_{n, \, 1 - \alpha}\\
0 &\mbox{else}
\end{cases}$$
where $\chi^{2-}_{n, \, 1 - \alpha}$ denotes the $1-\alpha$ quantile of the Chi-squared distribution with $n$ degrees of freedom. (I've heard that this is the uniformly most powerful test for our situation, is that correct?).
We now want to calculate the power of this test. I.e., given $X = (X_1, \dots, X_n)$, we evaluate for $\sigma^2 \in (0, \infty)$:
\begin{align}
\mathbb{E}_{\sigma^2}(\delta(X)) &= \mathbb{P}_{\sigma^2}(\delta(X) = 1)\\
&= \mathbb{P}_{\sigma^2} \left( \frac{1}{\sigma_0^2}\sum_{i=1}^n X_i^2 \geq \chi^{2-}_{n, \, 1 - \alpha} \right)\\
&= \mathbb{P}_{\sigma^2} \left( \frac{1}{\sigma^2}\sum_{i=1}^n X_i^2 \geq \frac{\sigma_0^2}{\sigma^2}\chi^{2-}_{n, \, 1 - \alpha} \right)\\
&= 1 - \mathbb{P}_{\sigma^2} \left( \frac{1}{\sigma^2}\sum_{i=1}^n X_i^2 \leq \frac{\sigma_0^2}{\sigma^2}\chi^{2-}_{n, \, 1 - \alpha} \right)
\end{align}
Note that $\left(\frac{1}{\sigma^2}\sum_{i=1}^n X_i^2 \right) \sim \chi^{2}_{n}$. Hence, when $\sigma^2 = \sigma_0^2$ we have
\begin{align}\mathbb{E}_{\sigma^2}(\delta(X)) &= 1 - \mathbb{P}_{\sigma^2} \left( \frac{1}{\sigma^2}\sum_{i=1}^n X_i^2 \leq \chi^{2-}_{n, \, 1 - \alpha} \right)\\
&= 1 - (1 - \alpha)\\
&= \alpha
\end{align}
My question: can we explicitly calculate $\mathbb{E}_{\sigma^2}(\delta(X))$ when $\sigma^2 \neq \sigma_0^2$?
 A: It seems you have verified
that when $H_0$ is true, the rejection rate of a
test at the 5% level is 0.05, as it should be.
Yes, it is possible to find the power of the test
for a particular alternative value of $\sigma_a^2.$
You can use an analytical method similar to what you did for
for the significance level when the null hypothesis
is true. (For numerical results you will need software or a suitable printed table of chi-squared distribution.)
I will show results from simulations. In many
power computations, simulation is necessary because
the distribution of the test statistic when $H_0$
is false is unknown or too messy to handle analytically. However, I hope you will see from
my simulations how to find the power analytically
for your question.
One test with null hypothesis true. With $H_0: \sigma^2 = 36$ true
tested against $H_a: \sigma^2 > 36$ at the 5% level, we do not reject because the P-value exceeds 5%.
For example, given a vector x of $n=100$ observations
from $\mathsf{Norm}(\mu = 0, \sigma=6),$ the test would
look like this:
set.seed(114)
n = 100;  sg = 6        
x = rnorm(n, 0, sg)           # data
v = mean(x^2);  v             # variance est
[1] 37.91495
h = n*v/36;  h                # test stat
[1] 105.3193
pv = 1 - pchisq(h,100);  pv   # P-val of test
[1] 0.3384785                 # > 5%; don't rej

Many tests. With null hypothesis true, testing at level 5%, the P-value should be less than
5% in 5% of the tests.
set.seed(2022)
pv = replicate(10^5, 
       1-pchisq(sum(rnorm(100,0,6)^2)/36, 100))
mean(pv <= .05)
[1] 0.05044         # aprx significance level
2*sd(pv <= .05)/sqrt(10^5)
[1] 0.001384143     # 95% margin of sim error

So, the significance level of the test is $0.0504 \pm 0.0014;$  essentially 5% as expected. (With 100,000
iterations we can expect about 2-place accuracy
for the result.)
Moreover, for an exact tests using a continuous-valued
test statistic, the distribution of P-values under $H_0$ is distributed $\mathsf{Unif}(0,1),$ as
illustrated by the histogram below. Rejection
occurs only for the 5% of instances to the left of
the vertical red line.
hist(pv, prob=T, col="skyblue2")
 abline(v = .05, lwd=2, col="red")


Many tests when the null hypothesis is false.
By contrast, when $\sigma^2 = 64 > \sigma_0^2 = 36,$
the rejection rate is much higher. We repeat
the simulation above with the alternative value
$\sigma_a^2 = 64.$
set.seed(2022)
pv = replicate(10^5, 
       1-pchisq(sum(rnorm(100,0,8)^2)/36, 100))
mean(pv <= .05)
[1] 0.99025    # aprx power of the test

The power of the test (rejection probability when
the null hypothesis is false in a particular way)
is about 99%.
Now, the distribution of the P-value is far from uniform.
hist(pv, prob=T, col="skyblue2")
 abline(v = .05, lwd=2, col="red")


