double integral with absolut value

[paul2834]

Hi,

So I tried to calculate this integral at region D in the image, by splitting it to two integrals.

I got the wrong answer at the end:

\(\displaystyle \displaystyle \int_D \int\ 2 |\, x\, |\, -\, \sqrt{y\,}\, dxdy,\, \). . .\(\displaystyle D\, :\, \left\{\, y\, =\, x^2,\, y\, =\, x\, +\, 2\,\right\}\)

I thought of splitting the integral into:

\(\displaystyle \displaystyle \int_{-1}^{0}\, dx\, \int_{x^2}^{x+2}\, -2x\, -\, \sqrt{y\,}\, dy\, +\, \int_0^2\, dx\, \int_{x^2}^{x+2}\, 2x\, -\, \sqrt{y\,}\, dy\)

Did I split it right?

 

[Ishuda]

Hi,

So I tried to calculate this integral at region D in the image, by splitting it to two integrals.

I got the wrong answer at the end:

\(\displaystyle \displaystyle \int_D \int\ 2 |\, x\, |\, -\, \sqrt{y\,}\, dxdy,\, \). . .\(\displaystyle D\, :\, \left\{\, y\, =\, x^2,\, y\, =\, x\, +\, 2\,\right\}\)

I thought of splitting the integral into:

\(\displaystyle \displaystyle \int_{-1}^{0}\, dx\, \int_{x^2}^{x+2}\, -2x\, -\, \sqrt{y\,}\, dy\, +\, \int_0^2\, dx\, \int_{x^2}^{x+2}\, 2x\, -\, \sqrt{y\,}\, dy\)

Did I split it right?

Since you have not show any work, I'm not sure. It does appear to me that you have split it correctly depending on just what you do with that square root of y. What is (x²)^3/2?

 

[paul2834]

Since you have not show any work, I'm not sure. It does appear to me that you have split it correctly depending on just what you do with that square root of y. What is (x²)^3/2?

(x²)^3/2? what do you mean? I didn't write something like that here...

 

[Deleted member 4993]

(x²)^3/2? what do you mean? I didn't write something like that here...

You didn't - but after integration, you should have a term like that!