# Is their a special way to factor when doing integrals

(Very sorry this could probably be cleared up if I had rights to comment, however I am new here) Anyways, I was going on stack overflow looking for help on a problem I keep getting wrong. For the record my class just went over these, it is very possible there are tricks we will learn in the future.

This thread covered the exact problem, except the user edited it later so it was revolving around the y axis not x axis. I need it revolved around the x axis.

You want the area between $\;y=\sqrt x\;,\;\;y=2\;$ and $\;x=0\;$ revolved around the $\;x$-axis, thus you get
$$\pi\int\limits_0^4\left(2^2-(\sqrt(x)^2\right)dx=\pi\left(16-\frac124^2\right)=8\pi$$
Just for fun and so I can try out the math scripting thin here is how I would tackle it. $$\pi\int\limits_0^4\left(2-\sqrt(x)\right)\left(2-\sqrt(x)\right)dx$$ $$\pi\int\limits_0^4\left(4-(2x)^2 + (1/4)x^2\right)dx$$ $$\pi\left[4x -\frac{((8x^\frac{2}{3}))}{2} + x\right]_0^4$$ $$=\pi\frac{8}{3}$$ If anyone sees my stupid mistake please let me know!
• To create a fraction $\frac{a}{b}$, use the command \frac{a}{b}. – Travis Jan 14 '16 at 6:50
• One thing: you expand $a^2-b^2$ as $(a-b)(a-b)$ while it should be $(a-b)(a+b)$. Easier, though, would be just to simplify $(\sqrt{x})^2$ to $x$ and evaluate the integral of $4-x$. – mickep Jan 14 '16 at 6:56