Question

Find the volume of the solid generated by revolving the standed region about the x-axis. Te volume of the solid is ? cubic units. The equation: 4x+3y=24

Find the volume of the solid generated by revolving the standed region about the x-axis.

Te volume of the solid is ? cubic units.
The equation: $$4x+3y=24$$

2021-05-20
Step 1
$$\Rightarrow y=8-\left(\frac{4}{3}\right)x$$
$$\Rightarrow y=4\times(2-\left(\frac{1}{3}\right)x)$$
for x intercept, $$y=0$$
$$0=4\times(2-\left(\frac{1}{3}\right)x)$$
$$\left(\frac{1}{3}\right)x=2$$
$$x=6$$
Volume generated by rotating the given region about x axis by washer method $$v=\int_{0}^{6}\pi[4\times(2-\left(\frac{1}{3}\right)x)]^{2}dx$$
$$v=\int_{0}^{6}16\pi[2^{2}-2\times2\left(\frac{1}{3}\right)x+(\left(\frac{1}{3}\right)x)^{2}]dx$$
$$v=\int_{0}^{6}16\pi[4-\left(\frac{4}{3}\right)x+\left(\frac{1}{9}\right)x^{2}]dx$$
$$v=\int_{0}^{6}16\pi[4x-\left(\frac{4}{3}\right)\left(\frac{1}{2}\right)x^{2}+\left(\frac{1}{9}\right)\left(\frac{1}{3}\right)x^{3}]$$
$$v=\int_{0}^{6}16\pi[4x-\left(\frac{2}{3}\right)x^{2}+\left(\frac{1}{27}\right)x^{3}]$$
$$v=16\pi[4\times6-\left(\frac{2}{3}\right)6^{2}+\left(\frac{1}{27}\right)6^{3}]-16\pi[4\times0-\left(\frac{2}{3}\right)0^{2}+\left(\frac{1}{27}\right)0^{3}]$$
$$v=16\pi[4\times6-\left(2\times\frac{36}{3}\right)+\left(\frac{216}{27}\right)]$$
$$v=16\pi[24-24+8]$$
$$v=128\pi$$
Volume $$=128\pi=402$$