/sci/ - Science & Math

[Anonymous No.9018172]

How is this supposed to be π/6 for a R=1 quarter sphere?
Am I a brainlet or did I buy a shitty book?

[Anonymous No.9018185]

I don't know what the bounds of integration are, but if you're only integrating over positive x and y, it's only one eighth of a sphere. Then, the volume is (1/8)*[(4/3)*pi*r^3] = pi/6.

[Anonymous No.9018191]

Since r is purely negative, you can replace r in the integral with -r to get an integral of r^2 over a quarter sphere and whatever the bounds of r are

[Anonymous No.9018192]

>>9018172
how did you describe the infitesimal surface element

[Anonymous No.9018193]

>>9018185
Right, I meant quarter hemisphere

[Anonymous No.9018195]

>>9018192
Standard Jacobian of polar coordinates

[Anonymous No.9018196]

>>9018172
Thanks, just needed to study for that

The book is right

[Anonymous No.9018219]

>>9018172
Calculate the integral for the whole sphere and divide it by 4

Just as the book says lets change it to polar coords, x = r*cos(theta) y=r*sin(theta)

As its a sphere we re in the region x^2+y^2 <= 1 , replacing terms you get r <= 1.

Therefore your integration limits are 0<=r<=1 and 0<=theta<=2pi

Now we just need to change the function inside of the integral to our new variables r and theta, say T is vector [x y] and T' the Jacobian derivating by r and theta, by variable change theorem the stuff inside is -> F(x,y) dxdy = F(r,theta)det(T')drdtheta. det(T') is r (calculate it)

So you have the limits of int, the new fuction (r*sqrt(1-r^2)) so just go ahead and integrate, the result is 2*pi/3, divided by 4, pi/6

[Anonymous No.9018244]

>>9018219
r =! R when integrating over r
I feel stupid now

[Anonymous No.9018253]

>>9018195
[math]2\pi\int_{0}^{1}\sqrt{1-r^2}dr = 2\pi[-(1-r^2)^{1/2}]_0^1=2\pi*[0-(-1)]=2\pi[/math]

got this

[Anonymous No.9018257]

>>9018253
it should be [math]rdr[/math]

[Anonymous No.9018258]

>>9018253
sorry messed up. i meant:

[math]2\pi\int_{0}^{1}\sqrt{1-r^2}rdr = 2\pi[-(1-r^2)^{1/2}]_0^1=2\pi*[0-(-1)]=2\pi[/math]

[Anonymous No.9018259]

>>9018253
>2\pi\int_{0}^{1}\sqrt{1-r^2}dr = 2\pi[-(1-r^2)^{1/2}]_0^1=2\pi*[0-(-1)]=2\pi
>got this

The mistake is this >>9018244

The function to integrate changes to 2\pi\int_{0}^{1}\r*sqrt{1-r^2}dr = 2\pi[-(1-r^2)^{1/2}]_0^1=2\pi*[0-(-1)]=2\pi

[Anonymous No.9018272]

>>9018258
You messed up the integral when doing the variable change of r

[Anonymous No.9018274]

>>9018272
more detail please

[Anonymous No.9018282]

>>9018274
The primitive is wrong, you have to do the variable change u = 1-r^2

[math]2\pi\int_{0}^{1}\sqrt{1-r^2}rdr = 2\pi[-\frac{1}{3 }(1-r^2)^{3/2}]_0^1[/math]

[Anonymous No.9018300]

>>9018282
you're right. thanks! I seem to not think straight at the locally late hour:

[math]2\pi\int_{0}^{1}\sqrt{1-r^2}rdr = 2\pi[-\frac{1}{3}(1-r^2)^{3/2}]_0^1=2\pi*[0-(-\frac{1}{3})]=\frac{2}{3}\pi[/math]

[Anonymous No.9019364]

>>9018300
You should only be integrating from pi/2 to 0 since the bounds are restricted to a quarter sphere