Find total charge (using double integration)

[Max]

The question asks to find total charge in a region given x has lower bound 0 - upper bound 5 , y has lower bound as 2 and upper bound as 5. Based on knowledge I have been reading throughout the chapter, I set up a double integration with those dxdy, but the results went out to be off - compared to the solution textbook that set up their double integration the other way around which was dydx.

My question is whether the order of dy and dx for double integration makes the difference when you attempt to find the volume?
Thanks!

 

[Mark44]

The question asks to find total charge in a region given x has lower bound 0 - upper bound 5 , y has lower bound as 2 and upper bound as 5. Based on knowledge I have been reading throughout the chapter, I set up a double integration with those dxdy, but the results went out to be off - compared to the solution textbook that set up their double integration the other way around which was dydx.

My question is whether the order of dy and dx for double integration makes the difference when you attempt to find the volume?
Thanks!

If the region of integration is the rectangle $\{(x, y) | 0 \le x \le 5, 2 \le y \le 5\}$, you should be able to integrate in either order. If the region is not a rectangle, then changing the order of integration makes a difference as far as the limits of integration go.

Since you haven't provided further information about the problem and your work, it's impossible to say why your answer disagrees with the answer in the book.

 

[Max]

Ah, thank you!

 

[Max]

And yes, today in lecture, the professor answered the same way as the region was a rectangle, so the boundaries inside should be in terms of the other variables while the boundary outside should be numbers

 

[Mark44]

And yes, today in lecture, the professor answered the same way as the region was a rectangle, so the boundaries inside should be in terms of the other variables while the boundary outside should be numbers

This makes no sense. If the region of integration is a rectangle, your integral will look something like this:
$\int_{x = 0}^5 \int_{y = 2}^5~f(x, y)dy~dx$.
Switching the order of integration gives $\int_{y = 2}^5 \int_{x = 0}^5~f(x, y)dx~dy$.
In both cases, the limits of integration don't involve variable expressions. I include "x = ..." and "y = ..." in the lower limits of integration only to show the name of the variable.

 

[Niladri Dan]

Will you please specify the question?