# Calc. Double Integrals: Setting Limits

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.

Related Introductory Physics Homework Help News on Phys.org
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: