# How do you calculate the double integral of #(xcos(x+y))dr# where r is the region: 0 less than or equal to x less then or equal to #(2pi)/6#, 0 less than or equal to y less than or equal to #(2pi)/4#?

(I've reduced the fractions.)

Now we need to find

OK, maybe that wasn't easiest. Now I have to do 2 integration by parts, but rather than re-starting, we'll keep on this path.

First Integral:

Second Integral:

Which evaluates to:

Adding the two integrals gives us:

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Could you clarify your question more? For one thing, you are sort of mixing rectangular and polar coordinates in the way you are stating it.

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To calculate the double integral of (x\cos(x+y)) over the region (r), where (0 \leq x \leq \frac{2\pi}{6}) and (0 \leq y \leq \frac{2\pi}{4}), you integrate with respect to (x) first and then with respect to (y).

The integral is evaluated as follows:

[ \int_{0}^{\frac{2\pi}{4}} \int_{0}^{\frac{2\pi}{6}} x\cos(x+y) ,dx,dy ]

First, integrate with respect to (x):

[ \int x\cos(x+y) ,dx = x\sin(x+y) - \int \sin(x+y) ,dx ]

[ = x\sin(x+y) + \cos(x+y) + C ]

Then evaluate the integral from (0) to (\frac{2\pi}{6}):

[ \left[x\sin(x+y) + \cos(x+y)\right]_{0}^{\frac{2\pi}{6}} ]

[ = \left[\left(\frac{2\pi}{6}\right)\sin\left(\frac{2\pi}{6}+y\right) + \cos\left(\frac{2\pi}{6}+y\right)\right] - \left[0\sin(0+y) + \cos(0+y)\right] ]

[ = \left(\frac{\pi}{3}\sin\left(\frac{\pi}{3}+y\right) + \cos\left(\frac{\pi}{3}+y\right)\right) - \cos(y) ]

Now, integrate the result with respect to (y) from (0) to (\frac{2\pi}{4}):

[ \int_{0}^{\frac{2\pi}{4}} \left(\frac{\pi}{3}\sin\left(\frac{\pi}{3}+y\right) + \cos\left(\frac{\pi}{3}+y\right) - \cos(y)\right) ,dy ]

[ = \left[-\frac{\pi}{3}\cos\left(\frac{\pi}{3}+y\right) + \sin\left(\frac{\pi}{3}+y\right) - \sin(y)\right]_{0}^{\frac{2\pi}{4}} ]

[ = \left(-\frac{\pi}{3}\cos\left(\frac{\pi}{3}+\frac{2\pi}{4}\right) + \sin\left(\frac{\pi}{3}+\frac{2\pi}{4}\right) - \sin\left(\frac{2\pi}{4}\right)\right) - \left(-\frac{\pi}{3}\cos\left(\frac{\pi}{3}\right) + \sin\left(\frac{\pi}{3}\right) - \sin(0)\right) ]

[ = \left(-\frac{\pi}{3}\cos\left(\frac{5\pi}{6}\right) + \sin\left(\frac{5\pi}{6}\right) - 0\right) - \left(-\frac{\pi}{3}\cos\left(\frac{\pi}{3}\right) + \sin\left(\frac{\pi}{3}\right) - 0\right) ]

[ = \left(-\frac{\pi}{3}\cdot\left(-\frac{\sqrt{3}}{2}\right) + \frac{1}{2} - 0\right) - \left(-\frac{\pi}{3}\cdot\left(\frac{1}{2}\right) + \frac{\sqrt{3}}{2} - 0\right) ]

[ = \frac{\pi\sqrt{3}}{6} + \frac{1}{2} - \frac{\pi}{6} - \frac{\sqrt{3}}{2} ]

[ = \frac{\pi\sqrt{3} + 3 - \pi\sqrt{3} - 3}{6} ]

[ = \frac{0}{6} ]

[ = 0 ]

Therefore, the double integral of (x\cos(x+y)) over the given region (r) is (0).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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