Aligning the shape so that it is axially symmetrical with respect to the z-axis, the volume of your shape will then by given by
<br />
V=\int_{0}^{2\pi}d\phi\int_{z_{l}}^{z_{u}}dz\int_{0}^{r\left( z\right)<br />
}rdr=\pi\int_{z_{l}}^{z_{u}}dzr^{2}\left( z\right) .<br />
Next we have to determine r\left( z\right). You've been given a radial profile [in a coordinate system \left( x,y\right) ], of a body that is axially aligned with the line y=x.
<br />
y=\ln\left( x\right) \text{, \ \ \ \ \ }4\leq x\leq10.<br />
Next we want to rotate our \left(x,y\right) coordinate frame so that the profile (the body rather) is axially alined with z. We do this by applying the transformation
<br />
z=\frac{x+y}{\sqrt{2}}\text{, \ \ \ \ \ \ \ \ }r=\frac{y-x}{\sqrt{2}}.<br />****Aside****
If you want to know where this transformation comes from then we can show it in the following way. If we map the curve into the complex plane by saying that
<br />
\zeta=x+iy<br />
(so that \operatorname{Re}\left(\zeta\right) =x and \operatorname{Im}\left(\zeta\right)=y) then we will plot out exactly the same curve as in the \left(x,y\right) plane, except now we're looking at the complex plane. This has the advantage that working with polar coordinates and rotations is simple. The function curve described by \zeta can be written as:
<br />
\zeta=r\left( x,y\right) e^{i\phi\left( x,y\right) }\text{.}<br />
Now if we want to rotate this curve by -\pi/4 [which turns the line y=x in the \left( x,y\right) coordinate system, onto the line \ y^{\prime}=0 in a new coordinate system, \left( x^{\prime},y^{\prime}\right)], then we simply multiply by e^{-i\pi/4}. This gives
<br />
\zeta^{\prime}=\zeta e^{-i\pi/4}=\zeta\frac{\left[ 1-i\right] }{\sqrt{2}<br />
}=\frac{\left[ x+iy\right] \left[ 1-i\right] }{\sqrt{2}}=\frac{x+y}<br />
{\sqrt{2}}+i\left( \frac{y-x}{\sqrt{2}}\right) .<br />
If we now take real and imaginary parts of \zeta^{\prime} our new basis becomes:
<br />
x^{\prime}=\frac{x+y}{\sqrt{2}}\text{, \ \ \ \ \ \ \ }y^{\prime}=\frac<br />
{y-x}{\sqrt{2}}.<br />****End Aside****
Parametrically the curve y=\ln\left(x\right) (in the \left(x,y\right) frame) can be written as
<br />
\nu\left( t\right) =\left( t,\ln\left( t\right) \right)<br />
or
<br />
x\left( t\right) =t\text{, \ \ \ \ \ \ \ \ \ \ \ \ \ \ }y\left( t\right)<br />
=\ln\left( t\right) \text{, }<br />
and the upper and lower limits for t are
<br />
t_{l}=4\text{, \ \ \ \ \ \ \ \ }t_{u}=10\text{.}<br />
Hence, in the rotated frame (the \left( r,z\right) frame), the parameterised curve is
<br />
z\left( t\right) =\frac{t+\ln\left( t\right) }{\sqrt{2}}\text{,<br />
\ \ \ \ \ \ \ \ }r\left( t\right) =\frac{\ln\left( t\right) -t}{\sqrt{2}}.<br />
Looking at the expression for the volume,
<br />
V=\pi\int_{z_{l}}^{z_{u}}dzr^{2}\left( z\right) ,<br />
since z is a function of t we can write
<br />
V=\pi\int_{t_{l}}^{t_{u}}dt\frac{dz}{dt}r^{2}\left( t\right) =\frac{\pi<br />
}{2\sqrt{2}}\int_{4}^{10}dt\left( 1+\frac{1}{t}\right) \left( \ln\left(<br />
t\right) -t\right) ^{2}=\frac{\pi}{2\sqrt{2}}\int_{\ln\left( 4\right)<br />
}^{\ln\left( 10\right) }\left( e^{s}+1\right) \left( s-e^{s}\right)<br />
^{2}ds<br />
<br />
=\frac{\pi\left(190.\,51\right) }{2\sqrt{2}}=211.6\left( \text{units of volume}\right) .<br />
Thankx!
