Evaluate the double integral?

Question

Evaluate the double integral from 0 to 1 and x^2 to 1 of (x^3)(sin(y^3)) dy dx by reversing the order of integration.

I know how to solve the integral, I just don't know how to set up the new integral, I don't know how to find the limits.

Answer 1

If you integrate over dy first, you can not get it done with analytical method. But if you reverse the order, then you can get it done very easily.

After switching the order, the original integral becomes

∫[0, 1] {∫[0, √y] (x^3)(sin(y^3)) dx}dy

= ∫[0, 1] (1/4)(y^2)(sin(y^3)) dy

= ∫[0, 1] (1/12)(sin(u)) du, u = y^3

= (-1/12)cos(u) from 0 to 1

= (1/12)(1 - cos(1)) <==Answer

Answer 2

To reverse the order of integration, you will need to draw a picture first. Notice that the original integral has x range from 0 to 1, thus the entire region must be sandwiched between the y-axis (x=0) and the line x=1. For each x, the y integral has lower bound of x² and an upper bound of 1. Thus the region is also sandwiched between the parabola y=x² and the line y=1.

Once you have drawn the picture, look at the overall lower and upper bounds for y in the region you drew. These should both be constants. Then for each y, you want to determine the lower and upper bounds for x. You can determine these bounds for x by drawing a horizontal line through the region. Hint: Solve y=x² for x in terms of y.

I hope this helps you.