Question

pls help me and no fake answer i really need help :(Show that the volume of the solid of revolution formed by revolving the region bounded by the graph of f(x) = (3x+5)^3, x =0 and x = 1 about the x-axis is 1/21 (8^7 -5^7) pi unit^3

Answer 1

Solution

Given the function

$f(x)=(3x+5)^3$

If we graph the equation, we would have.

when x = 0, f(x) = 5^3 = 125

when x = 1, f(x) = 8^3 = 512

Hence, the solid of revolution is given by

$V=\pi\int_0^1(3x+5)^6dx$

Evaluating the integral

$\begin{gathered} V=\frac{\pi}{7\times3}[(3x+5)^7]\text{ from 0 to 1} \\ \\ \end{gathered}$$\Rightarrow V=\frac{\pi}{21}(8^7-5^7)\text{ unit}^3$