Help double integral over a region question

[kidzonety]

I've tried this question with many different ways and i always got -11.576, but the autograder always marked it wrong. so hopefully i really did something wrong and you can teach me about it.

find the double integral of -3*x*y - 3*y over the region bounded by x^2 + y^2 = 9, y = 3x, and y = 0.

Thanks very much!

 

[tiny-tim]

Hi kidzonety!

(try using the X² tag just above the Reply box )

The obvious way is to convert to polar coordinates (though of course you could also do it in x and y) …

I expect you've already tried that, but anyway show us how you did it, and then we'll see what went wrong, and we'll know how to help!

 

[kidzonety]

Hi kidzonety!

(try using the X² tag just above the Reply box )

The obvious way is to convert to polar coordinates (though of course you could also do it in x and y) …

I expect you've already tried that, but anyway show us how you did it, and then we'll see what went wrong, and we'll know how to help!

Okay.
I just first converted the function to polar one which became -3*r³*cos(o)*sin(o) - 3*r²*sin(o) drdo
and just double integrate this over 0 <= r <= 3, and arctan(3) <= o <= pi/2.

 

[tiny-tim]

Hi kidzonety!

(have a theta: θ and an integral: ∫ and a ≤ )

Okay.
I just first converted the function to polar one which became -3*r³*cos(o)*sin(o) - 3*r²*sin(o) drdo
and just double integrate this over 0 <= r <= 3, and arctan(3) <= o <= pi/2.

ok (except it's 0 ≤ θ ≤ arctan(3)) … now split that into two integrals, ∫∫-3*r³*cosθ*sinθ drdθ and ∫∫-3*r²*sinθ drdθ,

which are now just ∫-3*r³dr*∫cosθ*sinθdθ and ∫-3*r²dr∫sinθdθ

 

[kidzonety]

Hi kidzonety!

(have a theta: θ and an integral: ∫ and a ≤ )


ok (except it's 0 ≤ θ ≤ arctan(3)) … now split that into two integrals, ∫∫-3*r³*cosθ*sinθ drdθ and ∫∫-3*r²*sinθ drdθ,

which are now just ∫-3*r³dr*∫cosθ*sinθdθ and ∫-3*r²dr∫sinθdθ

hey tiny-tim,
Ive found my mistake.
Thanks very much for your suggestions and advice!