This problem popped back up in the queue, so I had a look. (I'm assuming this was never completely resolved, to judge from OP's final comment on 13 Feb, 16:10.) In the first page image posted, the construction of the integral on line 5 is largely correct:
$$\int_{-r}^0 \ t \sqrt{r^2 - t^2} \ + \ R \sqrt{r^2 - t^2} \ \ dt \ . $$
But None has neglected the fact that since the equation for the circle is $ \ (x - R)^2 + y^2 = r^2 \ $ , which does not describe a function, the total "height" of the "vertical slices" runs from $ \ -\sqrt{r^2 - (x - R)^2} \ $ to $ \ +\sqrt{r^2 - (x - R)^2} \ $. The total volume integral for this "inner semi-torus" should thus be
$$V \ = \ 2 \cdot 2 \pi\int_{-r}^0 \ t \sqrt{r^2 - t^2} \ + \ R \sqrt{r^2 - t^2} \ \ dt \ , $$
after the first substitution. [EDIT: Now it's clear that None's EDIT 2 refers to this.]
The other issue is that the calculation of the first portion of the volume integral goes slightly astray. For the substitution $ \ u = r^2 - t^2 \ , \ du = -2t \ dt \ $ , the transformed limits are found correctly as $ \ u = 0 \ $ to $ \ u = r^2 \ $ . However, the replacement for $ \ t \ dt \ $ should have been $ \ -\frac{1}{2}\ du \ $ , making the result
$$\int_{-r}^0 \ t \sqrt{r^2 - t^2} \ dt \ \rightarrow \ -\frac{1}{2}\int_0^{r^2} \sqrt{u} \ du \ = \ -\frac{1}{3}r^3 ; $$
the other portion,
$$\int_{-r}^0 \ R \sqrt{r^2 - t^2} \ \ dt \ = \ \frac{\pi}{4} R \ r^2 $$
is determined correctly. Hence, the volume of this solid of revolution is
$$V \ = \ 4 \pi \cdot (\frac{\pi}{4} R \ r^2 \ - \ \frac{1}{3}r^3 ) \ = \ \pi^2 R \ r^2 \ - \ \frac{4 \pi}{3}r^3 \ . $$
(Assumingly, since the problem theme concerns consuming donuts, the second term suggests the volume of a spherical "donut hole".) This result is confirmed by the application of Pappus' (second) centroid theorem: a semi-circle of area $ \ \frac{1}{2} \pi \ r^2 \ $ is revolved around a circle of radius $ \ R - \frac{4}{3 \pi}r \ ^{*} $ , generating a volume of
$$ \ 2 \pi \cdot (R - \frac{4}{3 \pi}r) \cdot (\frac{1}{2} \pi \ r^2) \ = \ \pi^2 R \ r^2 \ - \ \frac{4 \pi}{3}r^3 \ . $$
$^{*}$ As the centroid of a semi-circle is a familiar result, I shan't derive it here.
To answer the question in the problem statement then, the remaining fraction of the original donut is given by
$$\frac{\pi^2 R \ r^2 - \frac{4 \pi}{3}r^3}{2 \pi^2 R \ r^2 } \ = \ \frac{1}{2} - (\frac{2}{3 \pi} \cdot \frac{r}{R}) \ , $$
so more than half of the donut has been eaten, as discussed in some of the comments.