What am I doing wrong with this integral?

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The integral being evaluated is ∫_{-1}^3 ∫_0^{3y} (x^2+y^2) dx dy. The initial attempt incorrectly equated parts of the integral, leading to confusion. A suggested approach is to split the integral into two separate parts: ∫_{-1}^3 ∫_0^{3y} x^2 dx dy and ∫_{-1}^3 ∫_0^{3y} y^2 dx dy. The correct evaluation for the first part involves recognizing that ∫_{-1}^3 x^2 dx from 0 to 3y results in (3y)^3/3. This clarification helps in correctly solving the integral.
duki
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Homework Statement



Homework Equations



\int_{-1}^3 \int_0^{3y} (x^2+y^2) dxdy

The Attempt at a Solution



I've gotten \int_{-1}^3 \frac{x^3}{3} = \int_{-1}^{3}y^3 dy for the first part, but this can't be right?
 
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How are you getting that equals sign in there?
Aren't you solving:
\int_{-1}^3 \int_0^{3y} (x^2+y^2) dxdy = ?

Try splitting it up to help you see it more easily, maybe:
\int_{-1}^3 \int_0^{3y}x^2dxdy+\int_{-1}^3 \int_0^{3y}y^2dxdy = ?
 
For the first integral I get:

\int_{-1}^3 y^3 dy
Is that right so far?
 
Well, no... because you know:

\int_{-1}^3 \int_0^{3y}x^2dxdy=\int_{-1}^3 \frac{(3y)^3}{3}dy
 
Oh. Oops. :)
Thanks
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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