# Double Integrals - Order of Integration

• Nov 24th 2009, 10:05 AM
billym
Double Integrals - Order of Integration
Find the area of A by evaluating the double integral

$\displaystyle \int\int_A \, dxdy$

where the region A is bounded by the curve $\displaystyle y=x^2-x$ and the line $\displaystyle y=2$

Is this correct:

$\displaystyle \int\int_A \, dxdy = \int_{-1}^{2}\int_{2}^{x^2-x} \, dydx$

Since it says evaluate $\displaystyle \int\int_A \, dxdy$, do you reckon I have to change the order of integration?

If so, I have:

$\displaystyle \int_{-1/4}^{2}\int_{1/2 \left(1-\sqrt{4y+1} \right)}^{1/2 \left(1+ \sqrt{4y+1} \right)} \, dxdy$

Is this ridiculous?
• Nov 24th 2009, 11:02 AM
tonio
Quote:

Originally Posted by billym
Find the area of A by evaluating the double integral

[tex]$\displaystyle \int\int_A \, dxdy$

where the region A is bounded by the curve $\displaystyle y=x^2-x$ and the line $\displaystyle y=2$

Is this correct:

$\displaystyle \int\int_A \, dxdy = \int_{-1}^{2}\int_{2}^{x^2-x} \, dydx$

Since it says evaluate $\displaystyle \int\int_A \, dxdy$, do you reckon I have to change the order of integration?

If so, I have:

$\displaystyle \int_{-1/4}^{2}\int_{1/2 \left(1-\sqrt{4y+1} \right)}^{1/2 \left(1+ \sqrt{4y+1} \right)} \, dxdy$

Is this ridiculous?

It looks pretty absurd: better, just change $\displaystyle dxdy$ by $\displaystyle dydx$ and do the first, much easier, integration.

Tonio