Consider the regions 
enclosed by the curves 
and 
. Set-up the integrals representing the following:
-
The perimeter of
.
-
The total area of the regions
and
.
-
The volume of the solid generated when
is rotated about the line
using the method of washers.
-
The volume of the solid generated when
is rotated about the line
using the method of cylindrical shells.
Solution
The regions are shown below.
The perimeter of consists of the horizontal segment of length 2, and the lengths of partial arcs of the two curves. We use the arc length formula to compute the perimeter 
. Since 
and 
, we have
Now, to compute the total area of the two regions, we consider horizontal strips:
In the region , the function at the right is 
and the function at the left is ![\displaystyle x=\sqrt[3]{y-\dfrac{1}{2}}](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+x%3D%5Csqrt%5B3%5D%7By-%5Cdfrac%7B1%7D%7B2%7D%7D&bg=ffffff&fg=333333&s=0 "\displaystyle x=\sqrt[3]{y-\dfrac{1}{2}}")
.
In the region , the function at the right is and the function at the left is .
Therefore, the total area is given by
![\displaystyle \mathcal{A} = \int_{-\frac{1}{2}}^{\frac{1}{2}}\left[-\left(y-\dfrac{1}{2}\right)^2-\sqrt[3]{y-\dfrac{1}{2}}\right]\,dy](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cmathcal%7BA%7D+%3D+%5Cint_%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5Cleft%5B-%5Cleft%28y-%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E2-%5Csqrt%5B3%5D%7By-%5Cdfrac%7B1%7D%7B2%7D%7D%5Cright%5D%5C%2Cdy&bg=ffffff&fg=333333&s=0 "\displaystyle \mathcal{A} = \int_{-\frac{1}{2}}^{\frac{1}{2}}\left[-\left(y-\dfrac{1}{2}\right)^2-\sqrt[3]{y-\dfrac{1}{2}}\right]\,dy")
![\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\mathcal{A}} + \int_{\frac{1}{2}}^{\frac{3}{2}}\left[\sqrt[3]{y-\dfrac{1}{2}}+\left(y-\dfrac{1}{2}\right)^2\right]\,dy](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Crule%7B0.0001em%7D%7B0.0001ex%7D+%5Cphantom%7B%5Cmathcal%7BA%7D%7D+%2B+%5Cint_%7B%5Cfrac%7B1%7D%7B2%7D%7D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Cleft%5B%5Csqrt%5B3%5D%7By-%5Cdfrac%7B1%7D%7B2%7D%7D%2B%5Cleft%28y-%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E2%5Cright%5D%5C%2Cdy&bg=ffffff&fg=333333&s=0 "\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\mathcal{A}} + \int_{\frac{1}{2}}^{\frac{3}{2}}\left[\sqrt[3]{y-\dfrac{1}{2}}+\left(y-\dfrac{1}{2}\right)^2\right]\,dy")
We now turn to the volumes of the two solids generated. Note that for number 3, we need representations for the outer and inner radii of the washers for the generated solid. But first, observe that we can express as 
in the region . Thus, the volume of the solid generated by revolving about 
is given by
![\displaystyle \mathcal{V}_{\mathcal{R}_{1}}=\pi\int_{-1}^{0}\left[ \left(x^{3}+\dfrac{1}{2}+\dfrac{1}{2}\right)^{2}-\left(\dfrac{1}{2}-\sqrt{-x}+\dfrac{1}{2}\right)^{2}\right]\,dx](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cmathcal%7BV%7D_%7B%5Cmathcal%7BR%7D_%7B1%7D%7D%3D%5Cpi%5Cint_%7B-1%7D%5E%7B0%7D%5Cleft%5B+%5Cleft%28x%5E%7B3%7D%2B%5Cdfrac%7B1%7D%7B2%7D%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B2%7D-%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D-%5Csqrt%7B-x%7D%2B%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7B2%7D%5Cright%5D%5C%2Cdx&bg=ffffff&fg=333333&s=0 "\displaystyle \mathcal{V}_{\mathcal{R}_{1}}=\pi\int_{-1}^{0}\left[ \left(x^{3}+\dfrac{1}{2}+\dfrac{1}{2}\right)^{2}-\left(\dfrac{1}{2}-\sqrt{-x}+\dfrac{1}{2}\right)^{2}\right]\,dx")
or
![\displaystyle \mathcal{V}_{\mathcal{R}_{1}}=\pi\int_{-1}^{0}\left[\left( x^{3}+1\right)^{2}-\left( 1-\sqrt{-x}\right)^{2}\right]\,dx.](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cmathcal%7BV%7D_%7B%5Cmathcal%7BR%7D_%7B1%7D%7D%3D%5Cpi%5Cint_%7B-1%7D%5E%7B0%7D%5Cleft%5B%5Cleft%28+x%5E%7B3%7D%2B1%5Cright%29%5E%7B2%7D-%5Cleft%28+1-%5Csqrt%7B-x%7D%5Cright%29%5E%7B2%7D%5Cright%5D%5C%2Cdx.&bg=ffffff&fg=333333&s=0 "\displaystyle \mathcal{V}_{\mathcal{R}_{1}}=\pi\int_{-1}^{0}\left[\left( x^{3}+1\right)^{2}-\left( 1-\sqrt{-x}\right)^{2}\right]\,dx.")
Next, to obtain the volume of the solid generated by rotating about , we note that the upper branch of can be expressed as 
in the region . Moreover, notice that the upper function varies in the region. Hence, by cylindrical shells, we express the volume as
![\displaystyle \mathcal{V}_{\mathcal{R}_{2}}=2\pi\int_{-1}^{0}\left(x+1\right)\left[\frac{3}{2}-\left(\frac{1}{2}+\sqrt{-x}\right)\right]\,dx](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cmathcal%7BV%7D_%7B%5Cmathcal%7BR%7D_%7B2%7D%7D%3D2%5Cpi%5Cint_%7B-1%7D%5E%7B0%7D%5Cleft%28x%2B1%5Cright%29%5Cleft%5B%5Cfrac%7B3%7D%7B2%7D-%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B-x%7D%5Cright%29%5Cright%5D%5C%2Cdx&bg=ffffff&fg=333333&s=0 "\displaystyle \mathcal{V}_{\mathcal{R}_{2}}=2\pi\int_{-1}^{0}\left(x+1\right)\left[\frac{3}{2}-\left(\frac{1}{2}+\sqrt{-x}\right)\right]\,dx")
![\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\mathcal{V}_{\mathcal{R}_{2}}}+2\pi\int_{0}^{1}(x+1)\left[\frac{3}{2}-\left(x^{3}+\frac{1}{2}\right)\right]\,dx](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Crule%7B0.0001em%7D%7B0.0001ex%7D+%5Cphantom%7B%5Cmathcal%7BV%7D_%7B%5Cmathcal%7BR%7D_%7B2%7D%7D%7D%2B2%5Cpi%5Cint_%7B0%7D%5E%7B1%7D%28x%2B1%29%5Cleft%5B%5Cfrac%7B3%7D%7B2%7D-%5Cleft%28x%5E%7B3%7D%2B%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Cright%5D%5C%2Cdx&bg=ffffff&fg=333333&s=0 "\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\mathcal{V}_{\mathcal{R}_{2}}}+2\pi\int_{0}^{1}(x+1)\left[\frac{3}{2}-\left(x^{3}+\frac{1}{2}\right)\right]\,dx")
or
![\displaystyle \mathcal{V}_{\mathcal{R}_{2}}=2\pi\left[\int_{-1}^{0}\left(x+1\right)\left(1-\sqrt{-x}\right)\,dx+\int_{0}^{1}(x+1)\left(1-x^{3}\right)\,dx\right].](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cmathcal%7BV%7D_%7B%5Cmathcal%7BR%7D_%7B2%7D%7D%3D2%5Cpi%5Cleft%5B%5Cint_%7B-1%7D%5E%7B0%7D%5Cleft%28x%2B1%5Cright%29%5Cleft%281-%5Csqrt%7B-x%7D%5Cright%29%5C%2Cdx%2B%5Cint_%7B0%7D%5E%7B1%7D%28x%2B1%29%5Cleft%281-x%5E%7B3%7D%5Cright%29%5C%2Cdx%5Cright%5D.&bg=ffffff&fg=333333&s=0 "\displaystyle \mathcal{V}_{\mathcal{R}_{2}}=2\pi\left[\int_{-1}^{0}\left(x+1\right)\left(1-\sqrt{-x}\right)\,dx+\int_{0}^{1}(x+1)\left(1-x^{3}\right)\,dx\right].")
This entry was posted on Sunday, March 23rd, 2008 at 4:48 pm and is filed under Calculus. You can follow any responses to this entry through the RSS 2.0 feed.
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March 23, 2008 at 10:16 pm |
The only thing more challenging than solving this particular problem would be explaining to me what the problem is even about. Sigh. Mathematics looks so beautiful and yet is such a complete mystery to me…
March 23, 2008 at 10:48 pm |
This equation, like all mathematical equations, is a lie. Something either happens, or it doesn’t
March 23, 2008 at 11:30 pm |
Thanks for clearing that up!
March 24, 2008 at 4:10 am |
i knew that.
March 24, 2008 at 7:40 am |
i think maths is complicated enough without new discoveries.