Help with integration problem.

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Homework Statement



find volume of y=4(x)^(.5) y=x, it revolves about x=17
here is my work
pi integral 0-16 (17-4(x)^(.5))^(2)-(17-x)^(2)
simplified and integrated too;
pi integral 0-16 {(50/2)x^2-(1/3)x^3-(272/3)(x)^(3/2)}
my definite integrated answer was
-768pi, but it is wrong

any help would be appreciated
thank you

Homework Equations


piR^2


The Attempt at a Solution

 
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I didn't check your work very thoroughly, but I can tell that you're missing a term in your second equation so you might want to recheck your expansion and integration steps. You should have a number times x in there if you wrote your first equation correctly. Also, you might want to look at the posts on how to use LaTeX. It makes your work a lot easier to read.
 
Are you trying to do it with the method of shells or discs? I couldn't follow your steps towards the solution. Try declaring your boundries/equations before starting on a solution.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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