Order Of Integration Switch

I've been given the double integral $\displaystyle \int_{x= 0}^{1}\int_{y=x}^{3x}xy dydx $ and instructed to change the order of integration and solve. After solving the given integrals I found they equaled 1.

I reasoned that the initial graph is bounded by two lines y=x and y=3x and bounded on the x-axis between x=0 and x=1; this formed a skewed sort of triangle.

Here's what I changed it to: $\displaystyle \int_{y= 0}^{3}\int_{x=(1/3)y}^{x}xy dxdy $

I either I'm I'm messing up or my professor's solution is wrong...help?

Re: Order Of Integration Switch

Did you just change around all the x's and y's? You must actually look at the geometry.

First Piece $\displaystyle \int_{0}^{1}\int_{y/3}^{y} xy\;dxdy$

Second Piece $\displaystyle \int_{1}^{3}\int_{y/3}^{1} xy\;dxdy$

Re: Order Of Integration Switch

When I solve the two integrals, I get 9 as opposed to 1 which was the answer prior to switching the order of integration. Should the two answers be the same?

Re: Order Of Integration Switch

Quote:

Originally Posted by

**spruancejr** When I solve the two integrals, I get 9 as opposed to 1 which was the answer prior to switching the order of integration. Should the two answers be the same?

Check your upper bound on the second set of integrals.

$\displaystyle \int_1^3 {\int_{y/3}^1 {xydxdy} } $

Re: Order Of Integration Switch

I'm confused by the second integral going from 0 to 1. Shouldn't it be from 1 to 3 since the second (top triangle) is bounded between 1 and 3 in the y direction?

Re: Order Of Integration Switch

Quote:

Originally Posted by

**spruancejr** I'm confused by the second integral going from 0 to 1. Shouldn't it be from 1 to 3 since the second (top triangle) is bounded between 1 and 3 in the y direction?

CORRECTED NOW.

That was a copy&paste error.