Answer to Question #222243 in Calculus for Besh

Question

The area under one arch of the sine curve revolves about the x-axis. Find the volume generated.

Answer 1

Solution;

Use the disc method;

$V=π\int_a^b(f(x))^2dx$

Given;

f(x)=sin(x)

For one arch of the sine curve function,the limits varies from 0 to π

a=0

b=π

Substitute in the formula;

$V=π\int_0^π(sin(x))^2dx$

$V=π\int_0^2(sin^2(x)dx$

By the trigonometric identity;

cos(2x)=1-2sin²(x)

Gives; $sin^2(x)=\frac{1-cos(2x)}{2}$

By substitution;

$V=π\int_0^π(\frac{1-cos(2x)}{2})dx$

$V=\fracπ2[\int_0^π(1-cos(2x))dx$

$V=\fracπ2[\int_0^π1dx-\int_0^πcos(2x)dx$

$V=\fracπ2[x]_0^π-\fracπ2[\frac{sin(2x)}{2}]_0^π$

$V=\fracπ2[π]-\fracπ4[sin(2π)-sin(0)]$

$V=\frac{π^2}{2}-\fracπ4(0)$

$V=\frac{π^2}{2}$

Answer;

$\frac{π^2}{2}$ cubic units.