<![CDATA[Math Is Fun Forum / Slicing method Integration Calculus 2.]]> 2020-09-15T16:19:37Z FluxBB http://www.mathisfunforum.com/viewtopic.php?id=25867 <![CDATA[Re: Slicing method Integration Calculus 2.]]> Hey thank you for the help, and you are correct it was the volume of revolution I got that exact setup and got 1/2 unit^3 but I forgot to add the pi like always thank you for the help!

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http://www.mathisfunforum.com/profile.php?id=222715 2020-09-15T16:19:37Z http://www.mathisfunforum.com/viewtopic.php?pid=415051#p415051
<![CDATA[Re: Slicing method Integration Calculus 2.]]> hi Double

Welcome to the forum.

Sketch the graph and draw two horizontal lines close together from the y axis to the curve, so you have drawn a typical strip.  This strip has area x across and dy width. If you were asked for the area of the part of the curve between y = 0 and y = 1 you would add up the strips like this:

You cannot integrate a function of x with respect to y, so you need to substitute with

The resulting function is directly integrable.

But you ask for a volume, so I'll assume this means that the area is rotated around the y axis to create a solid 'volume of revolution'.

Go back to the original strip and imagine it rotates around the y axis so as to generate a thin disc.  The area of this disc is π x^2 ( ie. pi radius squared ) and so its volume is π x^2 .dy  Here the new integral:

Once again you need to replace the function of x with one involving y .......    x^2 = 1 - y

So we then end up with this integral:

Can you finish from here?

Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2020-09-15T07:24:49Z http://www.mathisfunforum.com/viewtopic.php?pid=415041#p415041
<![CDATA[Re: Slicing method Integration Calculus 2.]]> it says to integrate and find volume

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http://www.mathisfunforum.com/profile.php?id=222715 2020-09-15T00:02:07Z http://www.mathisfunforum.com/viewtopic.php?pid=415036#p415036