Find volume using disk/washer/shell method

  • Thread starter Thread starter kari82
  • Start date Start date
  • Tags Tags
    Method Volume
kari82
Messages
34
Reaction score
0
Find volume of the solid generated revolving the region bounded by the graphs of the equations about line x=5

y=x, y=0, y=4, x=5

My plan is to use washer method. V=pi∫R(y)^2-r(y)^2 dy from y=0 and y=4

Im having trouble finding the equations for R and r. Can someone please explain me what would be a way to find those equations? Thanks!
 
Physics news on Phys.org
kari82 said:
Find volume of the solid generated revolving the region bounded by the graphs of the equations about line x=5

y=x, y=0, y=4, x=5

My plan is to use washer method. V=pi∫R(y)^2-r(y)^2 dy from y=0 and y=4

Im having trouble finding the equations for R and r. Can someone please explain me what would be a way to find those equations? Thanks!

Have you drawn a sketch of the region being revolved? The typical area element is a trapezoid whose left edge is along the line y = x and whose right edge is along the line x = 5.
 
yes, i did... and the book says that R=5-y and r=0.. but I can't see why r=0..
 
I think I got it.. r=0 because when it rotates along x=5 there is no inner radius?
 
Both radii are calculated as distances from the line x = 5. The large radius, R, is 5 - y. The small radius is 5 - 5 = 0. IOW, all of the disks have a radius of 5 - y. Does that make sense?
 
Yes! Thank you so much!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top