# Calc. Double Integrals: Setting Limits

1. Apr 2, 2005

### mkkrnfoo85

Here is the question:

Let D be the region bounded by the y axis and the parabola x = -4y^2 + 3. Compute 'double integral: (x^3)y dxdy'

I'm having a hard time setting limits to the double integral. I got y = [-sqrt(3/4), +sqrt(3/4)] for one of the limits, but I don't know how to set the x limits for integration. Also, can someone show me an easy way to quickly find the limits of integration. Thanks.

2. Apr 2, 2005

### whozum

For double integrals you always want the limits of the outer integral to be constants. Occasionally both integrals will have constant limits.

$$\int \int_D x^3y dx dy$$

The region D is contained by x = 0 and x = 3-4y^2. Since your outer integral is dy, your y limits will be constant, and x limits will be variable. Your limits alon the x axis are given by x = 0, and x = 3-4y^2, and:

$$\int \int_0^{3-4y^2} x^3y dx dy$$

Plot the graph of your domain to determine the range of y. You will see that the maximum y value occurs on the y axis, at x = 0.

Solving 0 = 3-4y^2, you find y = +/- sqrt(3/4).

The easiest way to find limits of integration is to sketch the region of integration.

The final integral becomes $$\int_{-\sqrt{3/4}}^{\sqrt{3/4}} \int_0^{3-4y^2} x^3y dx dy$$

Last edited: Apr 2, 2005