Tricky Integral

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ninguen

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Hi Guys,

First of all, thanks for your help. This site is awesome and I don't know what I'd do without you.

Now, on to the ugliness.

I have the following integral to solve: squiggle (x^3 - 4x^2 - 4x)^(1/2) dx

I've tried U substitution, but no dice. Too ugly.

As usual... Mathway says it can't be done. Schaum's says it can. The battle is on!

Chapter 30, This is problem 14.
 
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ninguen said:
I have the following integral to solve: squiggle (x^3 - 4x^2 - 4x)^(1/2) dx

Chapter 30, This is problem 14.
Click to expand...
That does not agree with my edition.
It appears that this is not a real integral.
 
Are you sure it is \(\displaystyle \displaystyle \int \sqrt{x^{3}-4x^{2}-4x}dx\)?.

If it were \(\displaystyle \displaystyle \int \sqrt{x^{3}-4x^{2}+4x}dx\), then we could work with it.

But, as is, it is a nasty Elliptic Integral and beyond regular calc methods.

If it were the latter, then we could write \(\displaystyle \int \sqrt{x(x-2)^{2}}dx\)
 
pka said:
That does not agree with my edition.
It appears that this is not a real integral.
Click to expand...

Consider the region R bounded by parabola y=(8x)^(1/2) and the line x = 2.

a) find the solid generated by revolving R around the y axis.

b) find the solid generated by revolving R around the line x = 2.

Maybe I'm screwing it up. I was doing part B.

Schaum's Calculus. Chapter 30. Question 14.
 
galactus said:
Are you sure it is \(\displaystyle \displaystyle \int \sqrt{x^{3}-4x^{2}-4x}dx\)?.

If it were \(\displaystyle \displaystyle \int \sqrt{x^{3}-4x^{2}+4x}dx\), then we could work with it.

But, as is, it is a nasty Elliptic Integral and beyond regular calc methods.

If it were the latter, then we could write \(\displaystyle \int \sqrt{x(x-2)^{2}}dx\)
Click to expand...

I think you're right. \(\displaystyle \displaystyle \int \sqrt{x^{3}-4x^{2}+4x}dx\)

I can take it from here.
 
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Calculate the volume by disc method - the integration will be much simpler.
 
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