Integrate Iterated Integrals: Solving Homework

[-EquinoX-]

Homework Statement

I need to evaluate this
$$\int_{-2}^{0} \! \int_{-\sqrt{9-x^2}}^{0} 2xy\,dy\,dx.$$

Homework Equations

The Attempt at a Solution

So I first tried integrating with respect to dy and I have:

$$xy^2$$ and evaluate that from $$-\sqrt{9-x^2}$$ to 0 , I will have $$-18x + 2x^3$$. Then I will integrate this again over dx, then I have $$9x^2 + \frac{x^4}{2}$$ and I evaluate this from -2 to 0. which results in -44. Where did I do wrong?

 

[Mark44]

Homework Statement

I need to evaluate this
$$\int_{-2}^{0} \! \int_{-\sqrt{9-x^2}}^{0} 2xy\,dy\,dx.$$

Homework Equations

The Attempt at a Solution

So I first tried integrating with respect to dy and I have:

$$xy^2$$

and evaluate that from $$-\sqrt{9-x^2}$$ to 0 , I will have $$-18x + 2x^3$$.

You should have gotten -x(9 - x^2) here, or x^3 - 9x. You're off by a factor of 2.

Then I will integrate this again over dx, then I have $$9x^2 + \frac{x^4}{2}$$ and I evaluate this from -2 to 0. which results in -44. Where did I do wrong?

It looks like you made another mistake, this time a sign error. From your previous work, you should have gotten -9x^2 + 1/2 * x^4.

The region over which integration takes place is a portion of the southwest quadrant of a circle of radius 3, centered at (0, 0). The region is bounded by the axes, the said circle, and the line x = -2.

Just to check my work, I integrated over the same sized region in the northeast quadrant of the circle, this time bounded by the line x = 2. You can't do this in general, but the integrand in this problem, 2xy, is unaffected by changes in signs to both variables. Both integrations resulted in values of 14, but the second way was simpler, with fewer chances to make sign errors.

 

[-EquinoX-]

So I should get 14 as a final result if I did everything correctly?

 

[Mark44]

Unless I made a mistake doing the problem in two different ways...

 

[HallsofIvy]

Yes, 14 is correct.