Why is this integration zero?

**nunos** · February 2nd 2010, 07:55 AM

Given the region $D = \{ (x,y): x^2-4 \leq y \leq 4 - x^2 \}$

Calculate the integral $\iint_D (1+y^2\sin y) dxdy$

In a solution I have for this exercise it 's said that $\iint_D (1+y^2\sin y) dxdy = \iint_D 1dxdy + \iint_D y^2\sin y dxdy$ and that $\iint_D y^2\sin y = 0$ .

My question is, why is that this integral equals to zero? (there's no explanation about that in the solution I have)

Thanks.

**HallsofIvy** · February 2nd 2010, 08:12 AM

Originally Posted by nunos

Given the region $D = \{ (x,y): x^2-4 \leq y \leq 4 - x^2 \}$

Calculate the integral $\iint_D (1+y^2\sin y) dxdy$

In a solution I have for this exercise it 's said that $\iint_D (1+y^2\sin y) dxdy = \iint_D 1dxdy + \iint_D y^2\sin y dxdy$ and that $\iint_D y^2\sin y = 0$ .

My question is, why is that this integral equals to zero? (there's no explanation about that in the solution I have)

Thanks.

sin y and so $y^2 sin y$ is an odd function. That is, sin(-y)= -sin(y) so $(-y)^2 sin(-y)= -y^2 sin(y)$ . The integral is from -4 to 4 and the integral from -4 to 0 cancels the integral from 0 to 4.

Keeping an eye on the "symmetry" of both function and region can simplify a lot of integrals!

**nunos** · February 2nd 2010, 08:24 AM

Thanks for your reply. However, there are some things I haven't quite got (yet, I hope).

1. Why is $y^2\sin y$ odd, only because $\sin y$ is odd?

2. Why does the integral from -4 to 0 cancels the integral from 0 to 4?

Thanks again.

**oblixps** · February 2nd 2010, 02:41 PM

the definition of an odd function is f(-x) = f(x). if you plug in -y into (y^2)siny, you get (y^2)sin(-y) and since sin is an odd function itself, sin(-y) = -sin(y) so (y^2)sin(-y) = -(y^2)siny and this shows that (y^2)siny is an odd function.

also from the definition of an odd function f(-x) = f(x), you can see that all odd functions are symmetric with respect to the origin. if you graph this function out notice that half the function is above the x axis and half is below the x axis. since you are integrating over a symmetric interval (from -4 to 4) both regions (from -4 to 0) and (from 0 to 4) are equal so when you add the two regions together you get zero. the integral is just a way to sum up an infinite number of objects so if you are adding negative objects to the positive objects of the same magnitude, they will cancel out to 0.

**nunos** · February 2nd 2010, 11:15 PM

Originally Posted by oblixps

the definition of an odd function is f(-x) = f(x). if you plug in -y into (y^2)siny, you get (y^2)sin(-y) and since sin is an odd function itself, sin(-y) = -sin(y) so (y^2)sin(-y) = -(y^2)siny and this shows that (y^2)siny is an odd function.

also from the definition of an odd function f(-x) = f(x), you can see that all odd functions are symmetric with respect to the origin. if you graph this function out notice that half the function is above the x axis and half is below the x axis. since you are integrating over a symmetric interval (from -4 to 4) both regions (from -4 to 0) and (from 0 to 4) are equal so when you add the two regions together you get zero. the integral is just a way to sum up an infinite number of objects so if you are adding negative objects to the positive objects of the same magnitude, they will cancel out to 0.

Does that mean, when I integrate over a function who has a line below the x-axis the integral is negative? Because if it's, it does make sense to me now, after your explanation... So, does it?

**HallsofIvy** · February 3rd 2010, 05:42 AM

Originally Posted by nunos

Does that mean, when I integrate over a function who has a line below the x-axis the integral is actually zero? Because if it's, it does make sense to me now, after your explanation... So, does it?

No, neither oblixps nor I said anything like that!

The function $x^2- 4$ goes below the x-axis and its integral from -2 to 2 is $\int_{-2}^2 x^2- 4 dx= \frac{1}{3}x^3- 4x\right|_{x=-2}^2= \frac{8}{3}- 8- (-\frac{8}{3}+ 8)= \frac{16}{3}- 16= -\frac{32}{3}$ .

What we are saying is that if a function has a section above the x-axis and a section below the x-axis, and they are both the same "size", then part above the x axis will be positive, the part below the x- axis will be negative and they will cancel.

That is very different from just "having a line below the x-axis". The parts above and below the x-axis must be exactly the same to cancel.

**nunos** · February 3rd 2010, 12:40 PM

Originally Posted by HallsofIvy

No, neither oblixps nor I said anything like that!

The function $x^2- 4$ goes below the x-axis and its integral from -2 to 2 is $\int_{-2}^2 x^2- 4 dx= \frac{1}{3}x^3- 4x\right|_{x=-2}^2= \frac{8}{3}- 8- (-\frac{8}{3}+ 8)= \frac{16}{3}- 16= -\frac{32}{3}$ .

What we are saying is that if a function has a section above the x-axis and a section below the x-axis, and they are both the same "size", then part above the x axis will be positive, the part below the x- axis will be negative and they will cancel.

That is very different from just "having a line below the x-axis". The parts above and below the x-axis must be exactly the same to cancel.

You are correct. And I think I am too. I just made the mistake of writing "below the x-axis is zero" instead of what I really meant, which was, "below the x-axis is negative".

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