Thread:

1. Double Integrals and Polar Coordinates

Hi there,

First off, I'm having a bit of trouble changing this integral's bounds so that I can evaluate it starting by integrating the dx part first. Any helps/tips will be appreciated.

$\int_1^3{\int_0^{\ln{x}} xdy}dx$

Second, I've just finished learning how to change variables into polar coordinates, but I'm still a bit confused on how to go about doing so. Here is an example I'm trying to evaluate, but I'm not quite sure where to start:

$\int_0^4{\int_{\sqrt{x}}^2{e}^{y^3}dy}dx$

Thanks.

2. Originally Posted by discretemather
Hi there,

First off, I'm having a bit of trouble changing this integral's bounds so that I can evaluate it starting by integrating the dx part first. Any helps/tips will be appreciated.

$\int_1^3{\int_0^{\ln{x}} xdy}dx$

Second, I've just finished learning how to change variables into polar coordinates, but I'm still a bit confused on how to go about doing so. Here is an example I'm trying to evaluate, but I'm not quite sure where to start:

$\int_0^4{\int_{\sqrt{x}}^2{e}^{y^3}dy}dx$

Thanks.
Why do you have to change the integration order in the first one?

And the second one you do not need to change to polar coordinates, just change the integration order.

3. Originally Posted by Mathstud28
Why do you have to change the integration order in the first one?

And the second one you do not need to change to polar coordinates, just change the integration order.
For the first one, I have to show that integrating it one way gives the same result as the other way.

For the second one, if changing the integration order, wouldn't I also have to change the bounds like in the first one?

4. Originally Posted by discretemather
Hi there,

First off, I'm having a bit of trouble changing this integral's bounds so that I can evaluate it starting by integrating the dx part first. Any helps/tips will be appreciated.

$\int_1^3{\int_{{\color{red}y = } 0}^{ {\color{red}y = }\ln{x}} xdy}dx$

[snip]
I find it helps to add the stuff in red. Did you try drawing the region that the integral terminals define? When you do, it should be evident that when you reverse the order of integration you have

$\int_{y=0}^{y=\ln 3} \int_{x=e^y}^{x=3} x \, dx \, dy$.

5. Originally Posted by discretemather
[snip]
Second, I've just finished learning how to change variables into polar coordinates, but I'm still a bit confused on how to go about doing so. Here is an example I'm trying to evaluate, but I'm not quite sure where to start:

$\int_0^4{\int_{{\color{red}y = }\sqrt{x}}^{{\color{red}y = }2} {e}^{y^3}dy}dx$

Thanks.
It's already been advised (correctly) to change the order of integration. And my previous remarks remain totally pertinent.

Upon reversal:

$\int_{y=0}^{y=2} \int_{x=0}^{x=y^2} e^{y^3} \, dx \, dy$.

6. While drawing the region of integration is the easiest,safest way out there, sometimes using inequalities will work too. If you have more algebraical affinities compared to geometrical, you can manipulate the inequalities.

The first bound says $0 \leq y\leq \ln x$

The second bound says $1 \leq x \leq 3$

Remember that both $\ln x$ and $e^x$ are increasing functions.

$1 \leq x \leq 3 \Rightarrow 0 \leq ln x \leq \ln 3$

$0 \leq ln x \leq \ln 3, 0 \leq y\leq \ln x \Rightarrow0 \leq y\leq \ln x \leq \ln 3$

Raise all sides to the power e.

$0 \leq y\leq \ln x \leq \ln 3 \Rightarrow 1 \leq e^y \leq x \leq 3$

So clearly $e^y \leq x \leq 3$ and $0 \leq y\leq\ln 3$

So the bounds on x are e^y and 3, while the bounds on y are 0 and ln 3

Thus $\int_1^3{\int_{0}^{\ln{x}} xdy}dx = \int_{0}^{\ln 3} \int_{e^y}^{3} x \, dx \, dy$

7. Thanks everyone! mr fantastic, for the first one, I had those exact bounds when I was trying it (I drew it out as well), but I guess I was making an error somewhere midway in the rest of my calculations (seems to work just fine on this try).

Isomorphism, that is an awesome way of figuring out the bounds! This will definitely help with my upcoming assignment. Thank you for sharing that!
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