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#### skatenerd

##### Active member

- Oct 3, 2012

- 114

This problem has brought something up that's making my brain wrinkle. It says to find the integral for

$$\int_{ }^{ }\int_{D}^{ }xy\,dA$$

where \(D\) is the region bounded by \(y=x\), \(y=2x-2\), \(y=0\). I have to find the \(dx\,dy\) integral and then find the \(dy\,dx\) integral and evaluate the simpler one.

I sketched the region which was easy enough and then found the easier of the two orders which was the \(dx\,dy\) integral and I got

$$\int_{0}^{2}\int_{y}^{\frac{y+2}{2}}xy\,dx\,dy$$

This evaluates to \(\frac{1}{6}\).

However I tried for the \(dy\,dx\) integral

$$\int_{0}^{2}\int_{2x-2}^{x}xy\,dy\,dx$$

and got the answer \(\frac{-38}{8}\) which definitely doesn't make sense. Where am I going wrong?

$$\int_{ }^{ }\int_{D}^{ }xy\,dA$$

where \(D\) is the region bounded by \(y=x\), \(y=2x-2\), \(y=0\). I have to find the \(dx\,dy\) integral and then find the \(dy\,dx\) integral and evaluate the simpler one.

I sketched the region which was easy enough and then found the easier of the two orders which was the \(dx\,dy\) integral and I got

$$\int_{0}^{2}\int_{y}^{\frac{y+2}{2}}xy\,dx\,dy$$

This evaluates to \(\frac{1}{6}\).

However I tried for the \(dy\,dx\) integral

$$\int_{0}^{2}\int_{2x-2}^{x}xy\,dy\,dx$$

and got the answer \(\frac{-38}{8}\) which definitely doesn't make sense. Where am I going wrong?

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