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

Answers (1)

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 \(=128\pi=402\)
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