! I kinda urgently need help with something: I've done this problem multiple times, and I know what the answer should be, but can't figure out what I'm doing wrong. Can anyone help me figure out what's wrong with my integration or simplifying? Thanks so much!


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
$$\begin{align*} \pi \int_{y = 1/10}^{100} \left(\sqrt{2y}\right)^2 - \left(\sqrt{2y} - \tfrac{1}{10}\right)^2 \, dy &= \pi \int_{y = 1/10}^{100} \tfrac{1}{10} \left( 2\sqrt{2y} - \tfrac{1}{10}\right) \, dy \\ &= \frac{\pi}{10} \left[ \frac{4 \sqrt{2}}{3} y^{3/2} - \frac{y}{10} \right]_{y=1/10}^{100} \\ &= \frac{\pi}{10} \left( \frac{4 \sqrt{2}}{3} (1000 - \tfrac{1}{10\sqrt{10}})- 10 + \frac{1}{100} \right). \end{align*}$$


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