- #1
Galadirith
- 107
- 0
Hi guys, I did post this over on the h/w c/w forum last week, but I think actually it is out of the scope of that, plus its not actually h/w c/w, its my general wondering :D
So I am having an issue with the volume of a sphere and calculating it. I am using the method where the solid, sphere in this case, is composed of many cylinders, in my case along the x-axis where the center of the sphere is at the origin.
So initially I attempted this by considering each cylinder was of volume:\pi y^2 \delta x
then the formula of the circle in the xy plane of the sphere is x^2 + y^2 = r^2 which yields y^2 = {r^2 - x^2} which means the volume of each cylinder becomes \pi (r^2 - x^2) \delta x
I will limit my range to [0,r] as it is symmetrical about the yz plane. So integrating accros the interval [0,r] we get :
V = \pi \int_{0}^{r} (r^2 - x^2) \delta x
= \pi ({r^2}x - \frac{x^3}{3})\bigg{|}_0^r
= \pi [({r^3} - \frac{r^3}{3})-({r^3}(0) - \frac{0^3}{3})]
= \pi \frac{2r^3}{3}
then x2 to get the volume of a sphere to be {4\pi r^3}{/}{3}
now that's all well and good, but I also wanted to see if another method would work, instead of using x as our independent use theta. Now I learned that if you wanted to calculate the surface are of a sphere you can use theta as it makes calculations easier. So using the idea that we can calculate the surface are of a sphere modeling it as composed of many hollow and cylinders with no base or head, each has height r\delta \theta. So each cylindrical surface becomes :
i. \ A = 2\pi y r \delta \theta
ii. \ y=rsin\theta
iii. \ A = 2r\pi rsin\theta \delta \theta
So that's fine, that idea can be taken as can successfully yield the surface area of a sphere successfully. Now I figured if that method can be employed for a surface, it must be adaptable to find the volume of a sphere. so this time I decided to define each solid cylinder as:
iv. \ V = \pi y^2 r \delta \theta
v. \ V = r\pi (r^2sin^2\theta) \delta \theta
so again I approached this as before, this time integrating across the range [0, \pi / 2] where I again will only find the hemisphere for +ve x and theta is the angle from the x axis. So my integration went like this :
V = \pi r^3\int_{0}^{\frac{\pi}{2}} (sin^2\theta) \delta \theta
= \pi r^3\int_{0}^{\frac{\pi}{2}} \frac{1}{2}(1 - cos2\theta) \delta \theta
= \frac{\pi r^3}{2}(\theta - \frac{1}{2}sin2\theta)\bigg{|}_{0}^{\frac{\pi}{2}}
= \frac{\pi r^3}{2}(\frac{\pi}{2}) = \frac{\pi^2 r^3}{4}
then oviously x2, but as is evident it is not the right answer, I really can't see what I have done wrong, I pretty sure none of my actually mathematics is wrong, which means the only part that could be wrong is the r\delta \theta bit and teacher suggested that for some reason it doesn't apply to a solid but does to a shell, but couldn't give a valid reason as to why. I am inclined to agree with him, but again I can't see why. In various texts I have looked though it says things along the lines of for the shell it is tempting to use \delta x but that r\delta \theta is a "better approximation", therefore surely it to is a better approximation for a solid of revolution to.
Can anyone explain why this is, what is the reason the integration can't be done this way, or am have I simply done my maths wrong, thanks guys :D
So I am having an issue with the volume of a sphere and calculating it. I am using the method where the solid, sphere in this case, is composed of many cylinders, in my case along the x-axis where the center of the sphere is at the origin.
So initially I attempted this by considering each cylinder was of volume:\pi y^2 \delta x
then the formula of the circle in the xy plane of the sphere is x^2 + y^2 = r^2 which yields y^2 = {r^2 - x^2} which means the volume of each cylinder becomes \pi (r^2 - x^2) \delta x
I will limit my range to [0,r] as it is symmetrical about the yz plane. So integrating accros the interval [0,r] we get :
V = \pi \int_{0}^{r} (r^2 - x^2) \delta x
= \pi ({r^2}x - \frac{x^3}{3})\bigg{|}_0^r
= \pi [({r^3} - \frac{r^3}{3})-({r^3}(0) - \frac{0^3}{3})]
= \pi \frac{2r^3}{3}
then x2 to get the volume of a sphere to be {4\pi r^3}{/}{3}
now that's all well and good, but I also wanted to see if another method would work, instead of using x as our independent use theta. Now I learned that if you wanted to calculate the surface are of a sphere you can use theta as it makes calculations easier. So using the idea that we can calculate the surface are of a sphere modeling it as composed of many hollow and cylinders with no base or head, each has height r\delta \theta. So each cylindrical surface becomes :
i. \ A = 2\pi y r \delta \theta
ii. \ y=rsin\theta
iii. \ A = 2r\pi rsin\theta \delta \theta
So that's fine, that idea can be taken as can successfully yield the surface area of a sphere successfully. Now I figured if that method can be employed for a surface, it must be adaptable to find the volume of a sphere. so this time I decided to define each solid cylinder as:
iv. \ V = \pi y^2 r \delta \theta
v. \ V = r\pi (r^2sin^2\theta) \delta \theta
so again I approached this as before, this time integrating across the range [0, \pi / 2] where I again will only find the hemisphere for +ve x and theta is the angle from the x axis. So my integration went like this :
V = \pi r^3\int_{0}^{\frac{\pi}{2}} (sin^2\theta) \delta \theta
= \pi r^3\int_{0}^{\frac{\pi}{2}} \frac{1}{2}(1 - cos2\theta) \delta \theta
= \frac{\pi r^3}{2}(\theta - \frac{1}{2}sin2\theta)\bigg{|}_{0}^{\frac{\pi}{2}}
= \frac{\pi r^3}{2}(\frac{\pi}{2}) = \frac{\pi^2 r^3}{4}
then oviously x2, but as is evident it is not the right answer, I really can't see what I have done wrong, I pretty sure none of my actually mathematics is wrong, which means the only part that could be wrong is the r\delta \theta bit and teacher suggested that for some reason it doesn't apply to a solid but does to a shell, but couldn't give a valid reason as to why. I am inclined to agree with him, but again I can't see why. In various texts I have looked though it says things along the lines of for the shell it is tempting to use \delta x but that r\delta \theta is a "better approximation", therefore surely it to is a better approximation for a solid of revolution to.
Can anyone explain why this is, what is the reason the integration can't be done this way, or am have I simply done my maths wrong, thanks guys :D