# 2 really advanced Calculus questions

• Dec 12th 2008, 02:15 PM
andreas
2 very difficult Calculus questions
Please, help somehow if you can!
• Dec 13th 2008, 04:33 AM
andreas
I did the following(for the first problem):

I imagined cylinder$\displaystyle x^2+y^2\leq1$ , $\displaystyle 0\leq Z\leq1$

First I consider the pipe without top and bottom ( $\displaystyle x^2+y^2=1$ and $\displaystyle 0<Z<1$ )

Since $\displaystyle F$ for $\displaystyle i$,$\displaystyle j$ has components 1 and 1. The integral for pipe will be simply $\displaystyle 2\pi$

Now I consider top and bottom discs: integral for bottom disc will be $\displaystyle 0$ because it's at $\displaystyle z=0$. The most interesting is top disc ( $\displaystyle x^2+y^2\leq 1$ and $\displaystyle Z=1$). For area of top disc I will have the following integral $\displaystyle \int z(x^2+y^2)^2 dA$ but the top z is 1 so integral is $\displaystyle \int (x^2+y^2)^2 dA$. dA is a little area at the top disc, and I think it is $\displaystyle 2\pi rdr$ where $\displaystyle r=\sqrt{x^2+y^2}$ So finally my integral is $\displaystyle \int (r^2)^2 2\pi rdr=2 \pi\int r^5 dr$ Taking it from 0 to 1 I get $\displaystyle \frac{\pi}{3}$. So the finall result is:$\displaystyle \frac{\pi}{3}+2\pi=\frac{7\pi}{3}$

I have to submit both problems very soon. Can you check if I did first correctly? How should I prove second one?
• Dec 13th 2008, 07:14 PM
mr fantastic
Quote:

Originally Posted by andreas
I did the following(for the first problem):

I imagined cylinder$\displaystyle x^2+y^2\leq1$ , $\displaystyle 0\leq Z\leq1$

First I consider the pipe without top and bottom ( $\displaystyle x^2+y^2=1$ and $\displaystyle 0<Z<1$ )

Since $\displaystyle F$ for $\displaystyle i$,$\displaystyle j$ has components 1 and 1. The integral for pipe will be simply $\displaystyle 2\pi$ Mr F says: I don't think so. See below.

Now I consider top and bottom discs: integral for bottom disc will be $\displaystyle 0$ because it's at $\displaystyle z=0$.The most interesting is top disc ( $\displaystyle x^2+y^2\leq 1$ and $\displaystyle Z=1$). For area of top disc I will have the following integral $\displaystyle \int z(x^2+y^2)^2 dA$ but the top z is 1 so integral is $\displaystyle \int (x^2+y^2)^2 dA$. dA is a little area at the top disc, and I think it is $\displaystyle 2\pi rdr$ where $\displaystyle r=\sqrt{x^2+y^2}$ So finally my integral is $\displaystyle \int (r^2)^2 2\pi rdr=2 \pi\int r^5 dr$ Taking it from 0 to 1 I get $\displaystyle \frac{\pi}{3}$.

So the finall result is:$\displaystyle \frac{\pi}{3}+2\pi=\frac{7\pi}{3}$

I have to submit both problems very soon. Can you check if I did first correctly? How should I prove second one?

I don't think the integral of F over the curved surface of the pipe is equal to $\displaystyle 2\pi$: If you parametrise the cylindrical surface then an infinitesimal element of the curved surface can be written as

$\displaystyle d\vec{S} = (\cos \theta \, i + \sin \theta \, j) (1) d\theta \, dz$.

Then the surface integral of F over the curved surface becomes

$\displaystyle \int_{z = 0}^1 \int_{\theta = 0}^{2 \pi} (\cos \theta + \sin \theta ) \, d \theta \, dz$

since $\displaystyle x^2 + y^2 = 1$ on the curved surface. I get zero as the answer.

So I think the integral over the closed surface will actually be $\displaystyle \frac{\pi}{3} + 0 = \frac{\pi}{3}$. Doing the calculation using the divergence theorem I get the same answer.