# How do you find the volume of the solid formed by revolving a particular region around the x-axis given #y=2#, #y=4-(x^2/2)# and bounded from [-2,2]?

This is a very good question indeed!!!!

First have a look at the region on the

Basically when this region rotates about the

To evaluate the volume you can first rotate the curved part alone and get its volume (a barrel) and then subtract the volume obtained rotating the straight line alone (a cylinder) as in:

To evaluate each volume I use a small cylinder of radius

So that:

And finally:

Hope it helps but PLEASE check my maths!

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To find the volume of the solid formed by revolving the given region around the x-axis, you would use the method of cylindrical shells. The formula for the volume using cylindrical shells is V = 2π∫(x)(f(x)) dx, where f(x) represents the height of the shell at a given x-value.

First, find the intersection points of the curves y = 2 and y = 4 - (x^2/2): 2 = 4 - (x^2/2) x^2/2 = 2 x^2 = 4 x = ±2

Now, integrate from -2 to 2 using the formula V = 2π∫(x)(f(x)) dx, where f(x) = 4 - (x^2/2) - 2: V = 2π∫(x)(4 - (x^2/2) - 2) dx from -2 to 2

After integrating, you'll find the volume of the solid formed by revolving the region around the x-axis.

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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.

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.

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.

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