Volume of solid of revolution.

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Dec 15th 2010, 10:01 PM
devil10078

Volume of solid of revolution.

1. Find the volume of the solid obtained by rotating the region bounded by the curve

y^2 - y^3 - x = 0, x = 0

about the line x = 0

2. Integral of (cos x + xsinx)/x(x+cosx) dx
Dec 15th 2010, 11:31 PM
FernandoRevilla

Quote:

Originally Posted by devil10078

1. Find the volume of the solid obtained by rotating the region bounded by the curve y^2 - y^3 - x = 0, x = 0 about the line x = 0

The intersection points are $(0,0)$ and $(0,1)$ . Then,

$V=\pi\displaystyle\int_0^1(y^2-y^3)^2\;dy=\ldots$

Fernando Revilla
Dec 16th 2010, 01:06 AM
Prove It

$\displaystyle \frac{\cos{x} + x\sin{x}}{x(x+\cos{x})}=-\left(\frac{-\cos{x} - x\sin{x}}{x^2 + x\cos{x}}\right)$

$\displaystyle = -\left(\frac{2x + \cos{x} - x\sin{x} - 2x- 2\cos{x} }{x^2 + x\cos{x}}\right)$

$\displaystyle = -\left(\frac{2x + \cos{x} - x\sin{x}}{x^2 + x\cos{x}}\right) - \left(\frac{-2x - 2\cos{x}}{x^2 + x\cos{x}}\right)$

$\displaystyle = -\left(\frac{2x + \cos{x} - x\sin{x}}{x^2 + x\cos{x}}\right) + \frac{2(x + \cos{x})}{x(x + \cos{x})}$

$\displaystyle = \frac{2}{x} - \frac{2x + \cos{x} - x\sin{x}}{x^2 + x\cos{x}}$ .

Therefore $\displaystyle \int{\frac{\cos{x} + x\sin{x}}{x(x + \cos{x})}\,dx} = \int{\frac{2}{x} - \frac{2x + \cos{x} - x\sin{x}}{x^2 + x\cos{x}}\,dx}$ .

To integrate the second term, make the substitution $\displaystyle u = x^2 + x\cos{x}$ .
Dec 17th 2010, 06:57 AM
Krizalid

good call, but it's actually easier than that, since

$\begin{aligned} \int{\frac{\cos x+x\sin x}{x(x+\cos x)}\,dx}&=\int{\frac{x+\cos x+x\sin x-x}{x(x+\cos x)}\,dx} \\ & =\ln \left| x \right|-\int{\frac{1-\sin x}{x+\cos x}\,dx} \\ & =\ln \left| x \right|-\ln \left| x+\cos x \right|+k \\ & =\ln \left| \frac{x}{x+\cos x} \right|+k. \end{aligned}$