Q. 51 - Calculus

Question

The region between the graph of $f\left(x\right)={(x-3)}^2+2$ and the x-axis on [0, 6], revolved around the y-axis

Step-by-Step Solution

Answer

The volume is $180\pi cubicunits$

Step 1: Given information

We are given a function as $f\left(x\right)={(x-3)}^2+2$

Step 2: Find the integral and evaluate it

We know that integral can be given as $V=2\pi {\int }_a^{b}r\left(x\right)h\left(x\right)dx$

The axis of revolution is y-axis hence the radius is $r\left(x\right)=x$

and the height can be given as $h\left(x\right)={(x-3)}^2+x$

Substituting the values in the formula we get,

$$\begin{gathered} V=2\pi {\int }_0^{6}x(x^2-6x+11)dx \\ V=2\pi {\int }_0^6(x^3-6x^2+11x)dx \\ V=2\pi {{[\frac{x^4}{4}-2x^3+\frac{11}{2}{x}^2]}^6}_0 \\ V=180\pi \end{gathered}$$