Q. 52 - Calculus

Question

The region between the graph of $f\left(x\right)=x^2$and the x-axis on [0, 3], revolved around the line y = 10

Step-by-Step Solution

Answer

The volume is $165.5\pi cubicunits$

Step 1: Given information

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

Step 2: Find the integral and evaluate it

The volume of the integral can be given as $V=2\pi {\int }_c^{d}r\left(y\right)h\left(y\right)dy$. The axis of revolution is y = 10

Hence the radius can be given as $r\left(y\right)=10-y$

And the height of the function can begiven as $h\left(y\right)=\sqrt{y}$

Substituting the values in the function we get,

$$\begin{gathered} V=2\pi {\int }_0^9(10-y)\left(\sqrt{y}\right)dy \\ V=2\pi {\int }_0^9(10\sqrt{y}-y\sqrt{y})dy \\ V=2\pi {{[\frac{20}{3}{y}^{\frac{3}{2}}-\frac{2}{5}{y}^{\frac{5}{2}}]}^9}_0 \\ V=2\pi [82.8] \\ V=165.5\pi \end{gathered}$$