![[../../00 Asset Bank/Pasted image 20240926150611.png|invert_B]] To calculate the [[Volume]] of a cylindricaly [[Symmetric]] shape, you can model the volume as a [[Definite Integral]] of a modified version of $f(x)$, where $f(x)$ is the height of the volume when $x$ away from the center. $\huge V = 2\pi \int_{a}^b xf(x)\d x $ >[!example] >Let $R$ be the region bounded by $y=-x+2$, $y=\sqrt{x}$ and $x=0$. $R$ is rotated about the $y$ [[Axis]] to form a solid shape $S$. Find the volume of $S$. >>[!check]- Solution >>$\huge\begin{align} -x+2 &= \sqrt{x} \\ (2-x)^2 &= x \\ x^2-4x+4&=x\\ x^2-5x+4&=0\\ (x-1)(x-4) &= 0\\ x &= \set{1, 4} \\ \\ h(x) &= 2-x-\sqrt x \\ V &= 2\pi\int_{0}^1 x\pa{2-x-\sqrt{x}} \d x \\ &= 2\pi \int_{0}^1 \pa{2x-x^2-x^{\frac{3}{2}}} \d x\\ &\cdots\\ &= \frac{8\pi}{15} \op{units}^3 \end{align} >>$ >[!example] >Let $R$ be the region bounded by $y=\sqrt{x-3}$, $y=0$ and $x=7$. $R$ is rotated about the $x$-[[Axis]] to form a solid $S$, find the [[Volume]] of $S$. >>[!check]- Solution >>$\huge \begin{align} >>y &= \sqrt{x-3} \\ >>x&= y^2 + 3 \\ \\ >>r(y) &= y \\ >>h(y) &= 7-(y^2+3) \\ >>&= 4-y^2\\ >>\\ >>V &= 2\pi \int_{0}^{\sqrt{7-3}} { r(y)h(y) }\d y\\ >>&= 2\pi \int_{0}^{2} y\pa{ 4-y^2 }\d y\\ >>&= 2\pi \int_{0}^{2} \pa{4y-y^3}\d y\\ >>&= 2\pi\pa{8-4}\\ >>&=8\pi \op{units}^3 >>\end{align}$