Volume by Shell and Washer Methods

  • Thread starter Thread starter rodneyspencer
  • Start date Start date
  • Tags Tags
    Shell Volume
rodneyspencer
Messages
3
Reaction score
0

Homework Statement



Find the volume generated by rotating the given region about the given line using the Shell method and the Washer method.

x = 4y [y = x/4], y = 0, x = 0, x = 8 about x



Homework Equations



Washer method (about x):
V = pi \int^b_a ((Rtop2(x) - rbottom2(x))dx

Shell method (about x):
V = 2pi \int^d_c (y[f(y)-g(y)])dy

The Attempt at a Solution



I'm not sure why I can't reconcile these two answers. I'm having some similar problems with more of these exercises but if someone can help me see where I'm going wrong I'm sure I can rework them successfully.

Washer:
V = pi \int^8_0 ((x/4)2)dx = 32pi/3

Shell:
V = 2pi \int^2_0 (y(4y))dy = 64pi/3
 

Attachments

  • Ex.jpg
    Ex.jpg
    5.2 KB · Views: 532
Last edited:
Physics news on Phys.org
I forgot about x=8. The shell method should be V = 2pi \int^2_0 (y(8-(4y)))dy = 32pi/3.

:facepalm:
 
Double-post :/

:facepalm:
 
Last edited:
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