Volume about x axis using cylindrical shells

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SUMMARY

The discussion focuses on calculating the volume of a solid obtained by rotating the region bounded by the curves \(x=1+y^2\), \(x=0\), \(y=1\), and \(y=2\) about the x-axis using the method of cylindrical shells. The correct volume is established as \(\frac{21\pi}{2}\). Participants noted that the method of washers is more effective for this problem, with a suggested formula of \(V = \pi \int_0^2 (2^2 - 1^2) \,dx + \pi \int_2^5 (2^2 - (\sqrt{x - 1})^2) \,dx\). The volume element is defined as \(dV=2\pi y(y^2+1) \,dy\), leading to the total volume calculation \(V=2\pi\int_1^2 (y^3+y) \,dy\).

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  • Familiarity with definite integrals and integration techniques
  • Knowledge of the equations of curves and their graphical representation
  • Ability to manipulate and simplify algebraic expressions
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karush
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Use the method of cylindrical shells to find the volume
of the solid obtained by rotating the region bounded by the given curves about the x axis.

$x=1+{y}^{2}$, $x=0$, $y-1$, $y=2$

https://www.physicsforums.com/attachments/4237

The answer is $\frac{21\pi}{2}$ but I couldn't get it using $V=2\pi\int_{a}^{b} \,xf(x)dx$
 
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karush said:
Use the method of cylindrical shells to find the volume
of the solid obtained by rotating the region bounded by the given curves about the x axis.

$x=1+{y}^{2}$, $x=0$, $y-1$, $y=2$
The answer is $\frac{21\pi}{2}$ but I couldn't get it using $V=2\pi\int_{a}^{b} \,xf(x)dx$

Washers work best in this case. Try $\displaystyle \begin{align*} V = \pi \int_0^2{ 2^2 - 1^2\,\mathrm{d}x } + \pi \int_2^5{ 2^2 - \left( \sqrt{x - 1} \right) ^2 \,\mathrm{d}x } \end{align*}$.
 
got it, I forgot to square the equations
 
karush said:
Use the method of cylindrical shells to find the volume...

The volume of an element is:

$$dV=2\pi y\left(y^2+1\right)\,dy$$

Thus, the total volume is:

$$V=2\pi\int_1^2 y^3+y\,dy$$

Computing the definite integral will give you the desired result. :D
 

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