Volumn by cross sections - solving in terms of which axis confusion?

[zeion]

Homework Statement

I need to find the volume of the solid generated by revolving this region bounded by the curves about the x-axis.

y = sqrt(x), x+y=6, y=1

Homework Equations

The Attempt at a Solution

I find the intersections of these curves I get:

(1,1), (4,2), (5,1)..
Then I see that to get the area of this cross section I need to do two sections from x = 1 to x = 4, and x = 4 to x = 5 if I integrated in terms of x..

So I see that it will be easier to integrate in terms of y:

I will get the area to be

[tex]
A(y) = \int_{1}^{2} [(-y+6)-(y^2)]dy
[/tex]

But now I am confused as to how I can formulate the integral to find the volume of this by revolving around the x-axis?

I know I need to use the difference of the boundary as the radius then multiply by pi.. but I don't understand how to do that if I wrote the area in terms of y?

 

[Mark44]

Forget the integral you have. That just gives you the area of the region, which isn't what you want.

Your typical volume element is a shell whose volume is 2* pi*radius*length*[itex]\Delta y[/itex]. For your problem radius is y, and length is (6 - y - y²). The limits of integration are the ones you found, y = 1 and y = 2.

BTW, you should have put this into the Calculus and Beyond section, not the Precalcus section.

 

[zeion]

So can I do this to find the volume?

[tex]

V = \int_{1}^{2} \pi[(-y+6)^2-(y^2)^2]dy

[/tex]

 

[Mark44]

So can I do this to find the volume?

[tex]

V = \int_{1}^{2} \pi[(-y+6)^2-(y^2)^2]dy

[/tex]

Reread what I wrote in post #2.