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Mathematics-Online problems:

Solution to the problem of the (previous) week

Problem:

Rotating the cross section shown below around the $ x$-axis produces a (not particularly stable) vase.



\includegraphics[width=\linewidth]{vase.eps}

$\displaystyle f(x)$ $\displaystyle =$ $\displaystyle \sqrt{x+1}$  
$\displaystyle g(x)$ $\displaystyle =$ $\displaystyle \sqrt{x-2}$  

To the nearest gram, what is the weight of the vase if it is made of glass having a density of $ 2.5\,$g/cm$ ^3$?


Answer:

$ m =$ g

Solution:

Mass of the solid of revolution:

$\displaystyle m$ $\displaystyle =$ $\displaystyle 2.5\pi\left(\int_0^{10}{f(x)^2}\; dx -\int_2^{10}{g(x)^2}\; dx\right) {\text g}$  
  $\displaystyle =$ $\displaystyle 2.5\pi\left(\int_0^{10}{x+1}\; dx -\int_2^{10}{x-2}\; dx\right) {\text g}$  
  $\displaystyle =$ $\displaystyle 2.5\pi\cdot28\: {\text g}$  
  $\displaystyle \approx$ $\displaystyle 220\: {\text g}$  

[problem of the week]