Applications of Integration-Volume

  • Thread starter Thread starter gusryder
  • Start date Start date
  • Tags Tags
    Applications
gusryder
Messages
2
Reaction score
0
I'm asked to find the volume generated by rotating the region bounded by the given curves about the y-axis (using the method of cylindrical shells).
I'm given the functions y= 4(x-2)^2and y = x^2 - 4x +7.
I'm not sure how to word this properly...they don't give me the domain of the function to find the volume...as in, most of the questions (and of course, all of the examples in the text) have given domains...find such and such when x=3 and x=0 or something of the like. I can do all the problems where the domain or range (in some cases) is given but I'm not entirely sure how to figure out my domains? Is that the intecept of the 2? Because then I would get x=1 and x=3. And if that's true then I'm just screwing something else up.
Another general question I have is when I'm making equations (and I'm consistently having this problem for areas etc), I always seem to subtract the wrong function from the other one...in other words, I always seem to end up with a negative or incorrect area/volume. Say for volumes...
\int_{a}^{b} 2 \pi xf(x)dx
for f(x) I always seem to subract the wrong function by the wrong function! How can I tell which one is going to be the correct one? And sometimes both answers are positive and one is correct and one is not. I asked my professor and he told us just to put a (+/-) at the front and then change it once you know what it is...?! At first I thought it was which function was "on top" of the graphed functions but that doesn't seem to work very well either!
Sorry for the super long post! It's been almost 4 years since I've done calc and now I have to take another course (calc II) so if my questions seem dumb, I'm sorry but I'm still trying to catch up! Thanks!
 
Physics news on Phys.org
for the domain, you are getting the right region
you set the functions equal to each other to find out where they intersect..
this may help
http://mathdemos.gcsu.edu/shellmethod/gallery/gallery.html
furthermmore,
the formula for the volume of a cylindrical shell is
V=2pi (delta r) h

so then
2*pi integral radius and your height
here radius is the distance from the y-axis to the center of the shell, which in this case is an x-distance..
now for the height, the height of the function is the top function minus the lower function...
hope this helps..
 
Last edited by a moderator:
Thanks again! Those animations are awesome!
 
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