hey guys i was working on a problem from my class and I'm having some problems with it. I keep getting a neg. answer which it can't be since I'm trying to find a volume of a function rotated around the y-axis
I'm posting this here hopefully i can get some pretty fast since i know alot of people are on this site.
here is a link to another forum that uses syntax to make it easier to see
problem with volume using shells : Calculus
the function being rotated is y=1/\sqrt{1-x^2}
boundaries are y=0 lower bound is x=0 and upper is x= sqrt{2}/2
the formula i used is y=2*\pi *r*h*w
r= radius and is x since function is rotated about y-axis
h=height and is Y(top)-Y(bottom) function is on top so it is just the function
so the set up is delta vol.=2*pi *x*(1/sqrt{1-x^2})delta x
so to integrate I need to do a U substitution my U I used is U=1-x^2
so du/dx=-2x
than subbing du in the integral becomes -\pi (1/sqrt{u})du
next i want to integrate with the u so than i would need to change the bounds
lower x=0 u=1-x^2 so u=1
upper x=sqrt{2}/2 u=1-x^2 so u=1/2
so my integral now looks like vol. = -pi int_{1}^{1/2}(1/sqrt{u})du
than integrating that i got 2*sqrt{u} so i can pull the two through integral
vol.= -2*pi \int_{1}^{1/2}(sqrt{u})du
now i want to solve it so i replace u with u=1-x^2 in intergral
this is where I'm having problems. i keep getting a neg. answer
what is have so far is vol.=-2*pi [ (sqrt{1-(1/2)^2})-(sqrt{1-1^2}) ]
vol.=-2*pi [\sqrt{1-(1/4)}-(\sqrt{1-1}) ]
-2*pi[ sqrt{3}/2-0 ]
thank you for any help,
-Tim
I'm posting this here hopefully i can get some pretty fast since i know alot of people are on this site.
here is a link to another forum that uses syntax to make it easier to see
problem with volume using shells : Calculus
the function being rotated is y=1/\sqrt{1-x^2}
boundaries are y=0 lower bound is x=0 and upper is x= sqrt{2}/2
the formula i used is y=2*\pi *r*h*w
r= radius and is x since function is rotated about y-axis
h=height and is Y(top)-Y(bottom) function is on top so it is just the function
so the set up is delta vol.=2*pi *x*(1/sqrt{1-x^2})delta x
so to integrate I need to do a U substitution my U I used is U=1-x^2
so du/dx=-2x
than subbing du in the integral becomes -\pi (1/sqrt{u})du
next i want to integrate with the u so than i would need to change the bounds
lower x=0 u=1-x^2 so u=1
upper x=sqrt{2}/2 u=1-x^2 so u=1/2
so my integral now looks like vol. = -pi int_{1}^{1/2}(1/sqrt{u})du
than integrating that i got 2*sqrt{u} so i can pull the two through integral
vol.= -2*pi \int_{1}^{1/2}(sqrt{u})du
now i want to solve it so i replace u with u=1-x^2 in intergral
this is where I'm having problems. i keep getting a neg. answer
what is have so far is vol.=-2*pi [ (sqrt{1-(1/2)^2})-(sqrt{1-1^2}) ]
vol.=-2*pi [\sqrt{1-(1/4)}-(\sqrt{1-1}) ]
-2*pi[ sqrt{3}/2-0 ]
thank you for any help,
-Tim