Volume of the solid of revolution

Hxzzl

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Feb 7, 2020
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I understand how to do solid revolution problems, but I keep ending up with decimals when it says it shouldn't.
The problem is below. Also, the answer I got is 9057.13.


The region bounded by f(x)= −4x^2+20x+56 , x=0 y=0 is rotated about the y-axis. Find the volume of the solid of revolution.
Find the exact value; write answer without decimals.
 
It doesn't say that you will get a whole number for an answer....it just says to write it that way.
 
It doesn't say that you will get a whole number for an answer....it just says to write it that way.
So if you do not get a whole number it says to write the answer as a whole number? Can you please show me where it says that?
 
Here is a diagram of the region to be revolved:

fmh_0121.png

Same as Jomo, I did not get your number as the volume of the solid. I chose to use the shell method for its relative simplicity.

Can you post your working so we can see what may have gone wrong?
 
Thanks for the help, but I figured it. Sorry I should've marked it solved.
 
I am probably misunderstanding this.

The exact value for the volume is:

[MATH]V=\frac{7546\pi}{3}[/MATH]
This is not a whole number, obviously. They just didn't want an approximation like:

[MATH]V\approx7902.152721329527[/MATH]
 
The exact value for the volume is:

[MATH]V=\frac{7546\pi}{3}[/MATH]
This is not a whole number, obviously. They just didn't want an approximation like:

[MATH]V\approx7902.152721329527[/MATH]
Ah. I see. So was I right-ish?
 
Ah. I see. So was I right-ish?

You were right when you said the problem does not say the result will be a whole number. But the problem just said to give the exact value, which has nothing to do with whole numbers.
 
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