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**Double****Member**- Registered: 2020-09-14
- Posts: 3

I'm having problems trying to figure this out please help.

the equation is y=1-X^2 and you must slice with squares and they must be perpendicular to the y-axis.

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**Double****Member**- Registered: 2020-09-14
- Posts: 3

it says to integrate and find volume

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,909

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

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Double****Member**- Registered: 2020-09-14
- Posts: 3

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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