Volumes by Integration (1 Viewer)

[dougal]

Find the volume of the solid formed when the line x + 3y - 1 = 0 is rotated about the x-axis from x = 0 to x = 8.

Also -

The area bound by the curve y = x^2 and the line y = x + 2 is rotated about the x axis. Find the exact volume of the solid formed.



I've been staring at volumes about the x and y axis all day long and they're beginning to blue before my eyes. Is there something simple & easy to remember for this? Especially with finding the volume of areas bound by two curves. I get confused because I find the area between the two curves, and then when it comes to finding the volume of that, I'm stumped.

 

[Riviet]

For the first one, rearrange your linear equation into y=1/3 - x/3. So your volume would be given by:

8
∫ π.y²dx=
0

8
∫ π(1/3 - x/3)²dx, expand it and then you can integrate it.
0

For the second one, find where the two curves meet by solving the two equations simultaneously:

x²=x+2
x²-x-2=0
(x-2)(x+1)=0
x=2,-1
We are only interested in the positive x-value because the area bounded by the two curves is in the first quadrant (helps if you draw a diagram).
Now we work out which curve is above the other, in this case it's the line y=x+2. So we basically integrate the difference of the square of the two functions. Therefore volume is given by:

2
∫ π[(y₁)² - (y₂)² dx]=
0
(Note: y₁ is the linear, y₂ is the quadratic)
2
∫ π[(x+2)²-(x²)² dx]
0

Expand this and you can integrate the rest.

 

[dougal]

Thanks.