Thread: Find the volume of the object using method of shells.

1. 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 .)

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

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

2. 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:

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

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

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

Check your result using the volume of a hemisphere formula. $V=\frac{2}{3}r^{3}$

3. 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 .)

$x=sqrt(4-y^2)$ 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 ...

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

if the whole sphere is required, double the result.

4. 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 - axis. (give your answer in terms of .)

$y= -x^2+13x-40$ 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.

5. Note that when we factor, we get $x^{2}-13x+40=(x-5)(x-8)$

This tells us where to break up the integration limits.

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

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