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Let R be the region between the graphs of #y=1# and #y=sinx# from x=0 to x=pi/2, how do you find the volume of region R revolved about the x-axis?

Answer 1

Aurora Alvarado

See the answer below:

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Answer 2

Paisley Johnson

To find the volume of region (R) revolved about the x-axis, you can use the method of cylindrical shells. The formula for finding the volume using cylindrical shells is:

[ V = 2\pi \int_{a}^{b} x \cdot f(x) , dx ]

Where:

(a) and (b) are the limits of integration, which in this case are (0) and (\frac{\pi}{2}).
(f(x)) is the height of the shell at the point (x), which is the difference between the upper function ((y = \sin(x))) and the lower function ((y = 1)).
(x) is the radius of the shell.

So, the volume of region (R) revolved about the x-axis can be found using:

[ V = 2\pi \int_{0}^{\frac{\pi}{2}} x \cdot (\sin(x) - 1) , dx ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.