# Find the volume of the object using method of shells.

• Nov 23rd 2009, 12:53 PM
matt1989
Find the volume of the object using method of shells.
Hello all.

My problem is this:

Rotate the region bounded by the given curves about the line indicated. Obtain the volume of the solid by the method of shells. (give your answer in terms of http://www.ilrn.com/ilrn/formulaImage?f=%5Cpi&ns=0.)

$\displaystyle x=sqrt(4-y^2)$ y-axis, x-axis. About the y-axis

I've been working on this for hours and cannot get it. Help would be appreciated.
• Nov 23rd 2009, 01:18 PM
galactus
When using shells, the cross sections are parallel to the axis we are revolving about. In this case, we are revolving about y so the cross sections will be parallel to the y-axis and 'stacked up' along the x-axis.

So, integrate in terms of x.

What we have here is a hemisphere of radius 2.

Solving the given equation for y in terms of x gives:

$\displaystyle y=\pm\sqrt{4-x^{2}}$

Since we are bounded by the axes, we are in the positive region:

$\displaystyle 2{\pi}\int_{0}^{2}x\sqrt{4-x^{2}}dx$

Check your result using the volume of a hemisphere formula. $\displaystyle V=\frac{2}{3}r^{3}$
• Nov 23rd 2009, 01:30 PM
skeeter
Quote:

Originally Posted by matt1989
Hello all.

My problem is this:

Rotate the region bounded by the given curves about the line indicated. Obtain the volume of the solid by the method of shells. (give your answer in terms of http://www.ilrn.com/ilrn/formulaImage?f=%5Cpi&ns=0.)

$\displaystyle x=sqrt(4-y^2)$ y-axis, x-axis. About the y-axis

I've been working on this for hours and cannot get it. Help would be appreciated.

note that the graph of this relation is a semicircle in quads I and IV

the solid formed by rotating the graphed region about the y-axis will yield a sphere.

what confuses me is the statement that the x-axis is a boundary for the region ... if so, then I suppose the region could either be the quarter circle in quad I or in quad IV.

if that is the case, the volume of the solid hemisphere formed by using cylindrical shells is ...

$\displaystyle V = 2\pi \int_0^2 x \sqrt{4 - x^2} \, dx$

if the whole sphere is required, double the result.
• Nov 23rd 2009, 01:59 PM
matt1989
First of all thanks for the quick replys that was really helpful and I was able to complete the problem!

Is it possible I could get help with 1 more problem? It is the same type of problem again but slightly different.

It is the following:

Use the method of shells to find the volume of the solid obtained by revolving the region bounded by the given curves about the http://www.ilrn.com/ilrn/formulaImag...%7D%5C%3A&ns=0- axis. (give your answer in terms of http://www.ilrn.com/ilrn/formulaImage?f=%5Cpi&ns=0.)

$\displaystyle y= -x^2+13x-40$ x=0 x=8 x-axis.

I know that I have to break it into a left half and right half and then add the two integrals together but I'm not sure if I have them set up correctly.

This problem seems extremly long the method that I've been going about it and I feel like I keep making mistakes some where. If I knew how to enter in all my work that I've done I would but I can't figure it out atm.

Any help would be appreciated.
• Nov 23rd 2009, 02:31 PM
galactus
Note that when we factor, we get $\displaystyle x^{2}-13x+40=(x-5)(x-8)$

This tells us where to break up the integration limits.

$\displaystyle 2{\pi}\int_{0}^{8}x|x^{3}-13x^{2}+40x|dx$

$\displaystyle 2{\pi}\left[\int_{0}^{5}x(x^{2}-13x+40)+\int_{5}^{8}x(-x^{2}+13x-40)\right]dx$