1. ## density

1. Find the area, mass, and center of mass of a thin plate bounded by $x^2-y^2 =1$ and $x = 4$ with density $\rho(x,y) = x$.

Computing these, should be OK. Just wondering about the limits of integration. Would it be $x = 0$ to $x = 4$ and $y = 0$ to $y = \sqrt{x^2-1}$?

2. Let $f(x,y) = \begin{cases} 1, \ \ \ \ \ \text{if} \ x \ \text{is irrational} \\ 4y^{3}, \ \ \text{if not} \end{cases}$

Show that $\int_{0}^{1} \left(\int_{0}^{1} f(x,y) \ dy \right) \ dx = 1$, but that $\int_{0}^{1} \left(\int_{0}^{1} f(x,y) \ dx \right) \ dy$ does not exist.

Then this implies that $\int_{0}^{1} f(x,y) \ dy = 1$ or $\int f(x,y) \ dy = y$. So we only integrate it when $f(x,y) = 1$? And for the second case, the function may be similar to $\int_{0}^{1} \frac{1}{x^2} \ dx$ which does not exist?

The only problem is, when we are integrating from $0$ to $1$, we integrate over both rationals and irrationals. So maybe the second case deals with this, and that is why it doesn't exist. Whereas, in the first case, we keep $x$ constant, and integrate over the rationals between $0$ and $1$?

2. Originally Posted by heathrowjohnny
1. Find the area, mass, and center of mass of a thin plate bounded by $x^2-y^2 =1$ and $x = 4$ with density $\rho(x,y) = x$.

Computing these, should be OK. Just wondering about the limits of integration. Would it be $x = 0$ to $x = 4$ and $y = 0$ to $y = \sqrt{x^2-1}$?
[snip]
No. Draw the graphs to see the region. Note that $
x^2-y^2 =1
$
is a hyperbola with centre at (0, 0) and vertices at (1, 0) and (-1, 0) .....

If first integrating with respect to y, the region is defined by $-\sqrt{x^2-1} \leq y \leq \sqrt{x^2-1}$ and $1 \leq x \leq 4$.

Alternatively, if first integrating with respect to x, the region is defined by $\sqrt{y^2+1} \leq x \leq 4$ and $-\sqrt{15} \leq y \leq \sqrt{15}$.

3. its not riemann integrable but it is lebesgue integral. so I dont think you can integrate first with $x$ (to be riemann integrable).

4. do you agree with me (i.e. its not riemann integrable but lebesgue integral)?

5. Originally Posted by heathrowjohnny
do you agree with me (i.e. its not riemann integrable but lebesgue integral)?
You're referring to question 2 ......? I haven't had a close look at it yet, but my off the cuff answer is yes, I do agree with you (I may regret saying that ....) Someone like TPH would be better qualified than me to comment on on this question ......