Your integration step on the last line of the first page is incorrect. You wrote $$\int_{y=0.1}^{100} \left( 0.2 \sqrt{2y} - 0.01 \right) \, dy = \left[ (0.2) \frac{2}{3} (2y)^{3/2} - 0.01y \right]_{y=0.1}^{100}.$$ This is not true; instead, since $$\sqrt{2y} = \sqrt{2} \sqrt{y}$$, you should have $$\int_{y=0.1}^{100} \left( 0.2 \sqrt{2y} - 0.01 \right) \, dy = \left[ (0.2) \sqrt{2} \frac{2}{3} y^{3/2} - 0.01y \right]_{y=0.1}^{100}.$$ This is also easily checked via differentiation of the antiderivative: $$\frac{d}{dy}\left[(0.2)\frac{2}{3}(2y)^{3/2}\right] = (0.2)(2y)^{1/2} \color{red}{\cdot \frac{d}{dy}[2y]},$$ by the chain rule. You neglected the factor in red; whereas $$\frac{d}{dy}\left[(0.2)\sqrt{2}\frac{2}{3} y^{3/2} \right] = (0.2)\sqrt{2} y^{1/2} = 0.2 \sqrt{2y}.$$
The entire computation can be made far simpler. I have elected to not use decimal notation. Using the difference of squares formula $$a^2 - b^2 = (a-b)(a+b)$$ with the choice $$a = \sqrt{2y}$$, $$b = \sqrt{2y} - \frac{1}{10}$$, we have $$a-b = \frac{1}{10}$$, $$a+b = 2\sqrt{2y} - \frac{1}{10}$$; hence
\begin{align*} \pi \int_{y = 1/10}^{100} \left(\sqrt{2y}\right)^2 - \left(\sqrt{2y} - \tfrac{1}{10}\right)^2 \, dy &= \pi \int_{y = 1/10}^{100} \tfrac{1}{10} \left( 2\sqrt{2y} - \tfrac{1}{10}\right) \, dy \\ &= \frac{\pi}{10} \left[ \frac{4 \sqrt{2}}{3} y^{3/2} - \frac{y}{10} \right]_{y=1/10}^{100} \\ &= \frac{\pi}{10} \left( \frac{4 \sqrt{2}}{3} (1000 - \tfrac{1}{10\sqrt{10}})- 10 + \frac{1}{100} \right). \end{align*}