Q. 55 - Calculus

Question

The region between the graph of $f\left(x\right)=9-x^2$ and the x-axis on [0, 3], revolved around the x-axis

Step-by-Step Solution

Answer

The volume is $23.8\pi cubicunits$

Step 1: Given information

We are given a function $f\left(x\right)=9-x^2$

Step 2: Find the integral and evaluate it

We know that integral can be given as $V=2\pi {\int }_c^{d}r\left(y\right)h\left(y\right)dy$

As the axis of revolution is x-axis we have $r\left(y\right)=y$

and $h\left(y\right)=\sqrt{9-y}$

Substituting the value in the integral as

$$\begin{gathered} V=2\pi {\int }_0^{3}y\sqrt{9-y}dy \\ V=2\pi [11.9] \\ V=23.8\pi \end{gathered}$$