Integrating with the TI-83 Calculator for Volume Approximation?

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To approximate the volume of the solid formed by revolving the region bounded by y = sin x and y = 0 from 0 to π about the y-axis, the integral setup is π ∫(0 to π) sin² x dx. To graph this on a TI-83, users should graph π sin² x and utilize the integral function found in the graph screen's options. The volume is derived from the area under the curve, calculated as πr², where r is the function value. An alternative method involves using the MATH menu to access the fnInt function for direct integration. Proper entry of the function and limits is crucial for accurate results.
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Homework Statement



Use the integration capabilities of a graphing utility to approximate the approximate the
volume of the solid formed by revolving the region bounded by the graphs of y = sin x and y 0 in the interval [0, \pi] about the y-axis. Round your answer to three decimal places.

Homework Equations



Yikes! Help!


The Attempt at a Solution



Can anyone help me set this up on a TI-83?
 
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First thing you need to do is setup the integral.

\pi \int_{0}^{\pi}{sin^2 x} dx

The integral gives you the area under the curve. So all you have to do is graph

{\pi}sin^2 x

Then on the graph screen go to the list with intersection/zero options and you should see an integral symbol. Select that then enter the limits 0 then pi
 
Ok, probably a silly question now, but how did you get from sinx to sin^2x?
 
Because it's pi*r^2

radius = r = the function your given.
 
I knew that.. long day.
 
Because volume = Area * length and area in this case is pi*r^2
 
Another way to do it on the calculator is to go to MATH and scroll down to option 9 which reads fnInt(. Select it then enter the function in, make sure you do it carefully, then press the comma button, X, and then the bounds of integration.

So according to how Feldoh set it up you would press MATH, 9, pi, sin, x, ), X^2, comma, X, comma, 0, comma, pi, )
 

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