Surface area and volume of a sphere

[vector22]

solve this integral

The area of a circle can be found by the washer method
The exact area of a washer is
$$dA = 2 \pi r \,\,dr \,\,\,\,\,\,\,eq.1$$
Area of a circle is:
$$\int 2 \pi r \,\,dr$$
Equation 1 mulitplied by h (height) gives the exact volume of a cylindrical shell:
$$dV = 2 \pi r h\,\,dr$$
From here it is possible to use trig functions to calculate the volume of a hemisphere by the method of cylindrical shells.
so
$$r = cos \theta$$
$$h = sin \theta$$
$$dr = -sin \theta \,\, d\theta$$

putting it all together

$$\int -sin^2 \theta \, cos \theta \,\, d\theta$$

Anyway, how do you solve that last integral??

 

[FeDeX_LaTeX]

Use the substitution $u = {\sin}\theta$. Do you see where to go from here?

 

[I like Serena]

Hi vector22!

Anyway, how do you solve that last integral??

What is the derivative of sin³theta?

 

[alexfloo]

Use the power rule, and the chain rule:

$\frac{d(\sin \theta)^3}{d \theta}(\sin \theta)^3 = \left(\frac{d}{d \sin \theta }(\sin \theta)^3\right)\cdot\left(\frac{d\sin \theta}{d \theta} \right)$

(Notice that if you cancelled, you'd end up with the original derivative.) If:

$\frac{d}{d \sin \theta }(sin \theta)^3$

looks weird then try substituting x:

$\frac{d}{d x}x^3$

and then after you compute the derivative (which is an easy one) you can substitute sin back in for x.

 

[vector22]

I like Serena - yep that is what the books say is the answer

I tried FeDeX_LaTeX suggestion u = sinθ to solve:
$$\int −sin^2\theta\, cos\theta \,\,d\theta$$
$$u = sin \theta \,\,\,\,\,\,\,\, u^2= sin^2\theta$$
$$du = cos \theta \,\,d\theta$$
rewrite original integral as:
$$\int u^2 \,\,\, du$$
$$= \frac{u^3}{3} \,\,\,$$
put the trig functions back in
$$= \frac{sin^3\theta}{3}$$
The exact volume of a cylindrical shell is: (from my first post)
$$dV=2\pi rh\,\,dr$$
The constant 2 pi, in above equation, is put outside the integration sign to calculate the volume of a hemisphere in the following definite integral:
$$V = 2\pi \int_0^{\pi/2} −sin2\theta \,cos\theta \,\,d\theta\,\,\,= \frac{2\pi sin^3\theta}{3} = \frac{2\pi}{3}$$

The fraction 2/3 means the hemisphere is 2/3 the volume of a cylinder of the same radius so the volume of a hemisphere is:

$$V = \frac{2\pi r^3}{3}$$
and sphere is twice that:
$$V = \frac{4\pi r^3}{3}$$

 

[vector22]

Solve this integral (surface area of hemisphere?)

So i was trying to come up with an exact surface area element of the sphere which I hope to be
$$dA = \frac {2\pi r}{\sqrt{1-r^2}}\,\, dr$$
This area element is much like the surface area of a spherical sector.


The surface area of the hemisphere should be:

$$A = 2\pi\,\int_0^1 \frac {r}{\sqrt{1-r^2}}\,\, dr$$
a little numerical analysis on the integral seems to indicate the answer is 2 pi but could be wrong.

Can anyone provide some details of how to solve that integral - yes superg33k I am looking at your post.

 

[superg33k]

u=1-r²

r=-1/2 du/dr

substitue these in

 

[RandomMystery]

Can you find the surface area of a sphere without using the cylinder formula?
(I mean, you can end up with the same formula, but you use different reasoning)

 

[superg33k]

$$A = 2\pi \int_0^1 \frac {r}{\sqrt{1-r^2}} dr = -\pi \int_1^0 \frac {1}{\sqrt{u}} \frac{du}{dr} dr = -\pi \int_1^0 \frac {1}{\sqrt{u}} du =$$

 

[vector22]

superg33k

The answer to that definite integral is 2pi is that what you get??
oh and why did you reverse the limits because the result I got was not negative with the limits left as is.

The "2" means the surface area of the hemisphere is twice the area of it's base circle so surface area of hemisphere is 2 pi r^2

Yes that is easily solved by the subst you suggested.

Random mystery yes I tried spherical coordinates and got the same answer.

 

[superg33k]

yeah i got 2pi and I reversed the limits because the limits were previously r=0 to r=1, and using u=1-r^2 then these same limits are u=1 to u=0.

 

[vector22]

Here are some more details

Solve the following indefinite integral and then plug in 2pi and the limits later on.
$$\int \frac{r}{\sqrt {1-r^2}} \,\,dr$$
substitute
$$u = 1-r^2 \,\,\,\,\,\,\,\,\,\,\,\, -\frac{1}{2}du = r\,\,dr$$
rewrite original integral with u as the new variable.
$$solve\,\,\,\,\, -\frac {1}{2} \int \frac{1}{\sqrt {u}} \,\,du$$
$$=-\frac {1}{2}\cdot \frac{2}{1} \sqrt {u}$$
replace u with 1-r^2
$$= -\sqrt {1-r^2}$$
so
$$2\pi\int_0^1 \frac{r}{\sqrt {1-r^2}} \,\,dr = 2\pi (-(\sqrt {1-1^2}) - (-\sqrt {1-0^2} ))$$
$$= 2\pi(0+1)\,\,\,\,\,\, = 2\pi$$
which is the surface area of the hemisphere only if the radius = 1
The surface area of the hemisphere is twice the area of it's base circle
$$2\pi r^2$$

 

[mathwonk]

as archimedes put it, a ball is a cone whose base is the surface sphere and whose height is the radius, hence the volume of a ball is R/3 times the surface area. so if you know the volume is 4/3 pi R^3 you are done.

moreover by cylindrical shells the volume of a ball equals 2piR/3 times the length of an interval of radius R, i.e. vol = (2piR/3)(2R) = (4/3)pi.R^3.

In general by the same argument, the volume of an n ball equals (2piR/n) vol(n-2 ball).

and the surface area of the n-ball is always the derivative wrt R of the volume. or if you prefer, the area is just n/R times the volume, by archimedes' remark in dimension n.

 

[vector22]

The integral can be used to find the surface area of a spherical cap

$$A = 2\pi r^2 \int_0^{a/r} \frac {x}{\sqrt{1-x^2}} \,\, dx$$

Where "a" is the radius of the spherical cap and "r" is the radius of the sphere. I changed the variable of integration from r to x to allow the 2 pi r^2 in which r is a constant. The functionality is exactly the same as before.

It also turns out that

$$r \int_0^{a/r} \frac {x}{\sqrt{1-x^2}} \,\, dx = \,\, h$$

where "h" is the height of the spherical cap, so the surface area of the spherical cap is also

$$A = 2\pi rh$$

 

[vector22]

I want to put all this in one post for reference. I started the ball rolling (no pun intended) with this thread.

https://www.physicsforums.com/showthread.php?t=521136
where I developed the idea of solving the volume of a sphere by trig functions.

Then in another thread, I solved for the surface area of a sphere (with some help) in this thread.

https://www.physicsforums.com/showthread.php?t=522836

Since my last post I would like to make it clear the connection between the surface area and volume of a sphere. (Mathwonk touched on this before)

The surface area of a sphere is:
$$4\pi r^2$$
The exact volume of a spherical shell is:
$$dV = 4\pi r^2 \,\, dr$$
Integrate to find the volume of a sphere
$$V = 4\pi \int r^2 \,\, dr$$
$$= \frac{4\pi r^3}{3}$$

 

[berkeman]

(Moderator's Note -- 3 separate threads merged into this one).