Question

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

Answer 1

so we have \[\int^{1}_{0}\int^{1}_{x^2}x^3sin(y^3)dydx\]correct?

Answer 2

are you familiar with Fubini's theorem?

Answer 3

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

Answer 4

Ok so what we want to do is figure out what our bounds on y actually are

Answer 5

How so? What's the fastest way to find out without graphing?

Answer 6

well our second integration in this case is the dx, so we know that our x is constrained as such: 0≤x≤1

Answer 7

so what is the smallest x^2 can be?

Answer 8

1?

Answer 9

I mean, 0?

Answer 10

yep why?

Answer 11

Because I plugged in 0 in x^2.

Answer 12

ok good so other than graphing, which is useful when you are eprmitted a calculator, but too time consuming when you do not, you have to use logic. If I show you 0≤x≤1 and x^2≤y≤1 can you just observe and see what the bounds are by thinking?

Answer 13

Wait a minute, let me work it out.

Answer 14

The new limits would be 0<=y<=1, x^2<=x<=1, right?

Answer 15

not quite you can't have 2 x's in the same inequality try again

Answer 16

focus on the second inequality I typed. Could you maybe take the square root of each term?

Answer 17

So it'd be x<=sqrt(x)<=1, right?

Answer 18

no remember you cannot have two x's in the same inequality. why? because then you are left with a variable post integration that you can't do anything with.

Answer 19

maybe thinking of it as we want the bounds on x in terms of y will help

Answer 20

So it's sqrt(y)<=x<=1, right?

Answer 21

not quite look at this inequality and tell me if √y is ≤ or ≥ x :
x^2≤y≤1