Double Int.

Tutors: The image quality is rather poor. The integral and working appears to be as follows:

\(\displaystyle \L \int_{x=0}^2 \, \int_{y=x}^2 \, 2\, y^2\, \sin{(xy)}\, dy\,dx\)

Instructions: Reverse order of integrate and [illegible].

Code: 
       /|\
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      2 ------* (2, 2)
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x       |     2
        |y
\(\displaystyle \L \int_{y=0}^2\, \int_{x=0}^2\, 2\, y^2\, \sin{(xy)}\, dx\, dy\)

. . . . . . . . . .\(\displaystyle \L =\, \int_{y=0}^2\, \left[\, 2\, x^2\, y\, (-\cos{(xy)})\, \right]\, |_{x=0}^2\, dy\)

. . . . . . . . . .\(\displaystyle \L =\, \int_{y=0}^2\, \left(\, 4\, y^2\, (-\cos{(2y)})\, \right)\, dy\)

. . . . . . . . . .\(\displaystyle \L =\, \left[\, \left(\frac{4}{3}\right)\, y^3\, (-\sin{(2y)})\, \right]\, |_0^2\)

. . . . . . . . . .\(\displaystyle \L =\, \left[\, \left(\frac{4}{3}\right)\, (8)\, (-\sin{(4)})\, \right]\)

. . . . . . . . . .\(\displaystyle \L =\, \left(\frac{32}{3}\right)\, (-\sin{(4)})\)
Click to expand...
 
Cove-Creek said:
reverse and eval.

What have I done wrong?

http://www.hunt101.com/img/486288.jpg
Click to expand...
Notice that your new integral limits represent a square, contrary to your correct diagram. On your diagram, you want to consider horizontal strips. From left to right, for each strip we have x going from 0 (the y-axis) to y (the diagonal line). Such strips are summed from y=0 (the x-axis) to y=2.

Moreover, your integrating is a bit mixed up. \(\displaystyle \mbox{ \int 2y^2sin(xy) dx = -2y\cos{(xy)}}\), for example.
 
Sorry for the poor image quality.

I am following you when you say to look at the horizontal strips but I am still not clear on what my new integral limits should be.
 
limits for dx should be x = 0 to x = y and limits for dy should be y = 0 to y = 2

I think that I understand it now, thanks unco
 
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