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How do you find the volume bounded by #y=sqrt(x + 1)#, x = 0, x = 3, and y = 0 revolved about the x-axis?

Answer 1

Luke Alvarez

We will use the disk method. Where #r=y#

The volume is #V=int_0^3pir^2dx=int_0^3pi(sqrt(x+1))^2dx=# #piint_0^3(x+1)dx=pi[x^2/2+x]_0^3=pi(9/2+3)=(15pi)/2#

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Answer 2

Emma Aldridge

To find the volume bounded by ( y = \sqrt{x + 1} ), ( x = 0 ), ( x = 3 ), and ( y = 0 ) revolved about the x-axis, you can use the disk method or washer method. Since the region is bounded by the x-axis and the curve, you'll integrate with respect to ( x ). Calculate the integral from ( x = 0 ) to ( x = 3 ) of ( \pi (\text{radius})^2 ) where the radius is the distance from the curve to the x-axis. The formula for the radius is ( y ) in this case. Therefore, the integral becomes ( \pi \int_{0}^{3} (\sqrt{x + 1})^2 , dx ). Integrating this expression will give you the volume of the solid generated by revolving the region about the x-axis.

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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.