Thread: Double and triple integral issues

I'm having a really hard time trying to figure out how to find the limits of double and triple integrals.

I'm also having trouble determining which limits go on which integral and whether it should be ordered or , etc.

Here's a few problems from the book:

Find the area of the region R that lies below the parabola y=4x-x^2 above the x-axis, and above the line y=-3x+6.

It shows that area = (I'm not sure what's wrong with the LaTeX? here's what I used: \int_{1}^{2} \int_{-3x+6}^{4x-x^2} dydx + \int_{2}^{4} \int_{0}^{4x-x^2} dydx )

No idea how they got those limits.

Originally Posted by dataspot

I'm having a really hard time trying to figure out how to find the limits of double and triple integrals.

I'm also having trouble determining which limits go on which integral and whether it should be ordered or , etc.

Here's a few problems from the book:

Find the area of the region R that lies below the parabola y=4x-x^2 above the x-axis, and above the line y=-3x+6.

It shows that area = (I'm not sure what's wrong with the LaTeX? here's what I used: \int_{1}^{2} \int_{-3x+6}^{4x-x^2} dydx + \int_{2}^{4} \int_{0}^{4x-x^2} dydx )

No idea how they got those limits.

Have you drawn a careful sketch of the region? Note that one of the intersection points of the line and the parabola is (1, 3):

-3x + 6 = 4x - x^2 => x^2 - 7x + 6 = 0 => (x - 6)(x - 1) = 0 => x = 1, 6.

If you integrate first with respect to y, then there are two distinct subregions:

R1: From y = -3x + 6 to y = 4x - x^2. And note that x goes from 1 to 2. Set up the integral ......

R2: From y = 0 (the x-axis) to y = 4x - x^2. And note that x goes from 2 to 4. Set up the integral ......