Sun K.

asked • 05/04/13# Evaluate the double integral?

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.

## 2 Answers By Expert Tutors

Robert J. answered • 05/04/13

Certified High School AP Calculus and Physics Teacher

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

Roman C. answered • 05/04/13

Masters of Education Graduate with Mathematics Expertise

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 x^{2} and an upper bound of 1. Thus the region is also sandwiched between the parabola y=x^{2} 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=x^{2} for x in terms of y.

I hope this helps you.

## Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem.

Most questions answered within 4 hours.

#### OR

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.

Sun K.

So the new limits would be 0<=y<=1 and sqrt(y)<=x<=1, right?

05/04/13