MHB Double integral of general region

[skate_nerd]

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?

 

[Prove It]

Re: double integral of general region

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?

Your bounds are correct, so it must be with your integration. What did you do?

 

[chisigma]

Re: double integral of general region

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?

On the basis of geometrical evidence the integral has to be splitted into two integrals... $\displaystyle \int \int_{D} x\ y\ dA = \int_{0}^{1} \int_{0}^{x} x\ y\ d y\ d x + \int_{1}^{2} \int_{0}^{2x-2} x\ y\ d y\ d x$ (1)... and the integration is left to You...

Kind regards

$\chi$ $\sigma$

 

[skate_nerd]

Chisigma, are you positive that this is the right way to do it? Because the problem states:
a) Set up the integral for \(\int_{ }^{ }\int_{D}^{ }xy\,dA\) as a \(dydx\) integral
b) Set up the integral for \(\int_{ }^{ }\int_{D}^{ }xy\,dA\) as a \(dxdy\) integral
c) Solve the simpler of the two integrals
So is the wording of this problem just meant to be tricky, since its actually going to be two integrals for each order of \(\int_{ }^{ }\int_{D}^{ }xy\,dA\)?

 

[skate_nerd]

Oh and @Prove It, turns out I did indeed mess up both of those integrals (Doh) however I am positive I got the right answers now and they are still different from each other. I got \(\frac{5}{6}\) for the \(dx\,dy\) integral and \(\frac{2}{3}\) for the \(dy\,dx\) integral.

 

[skate_nerd]

Actually Chisigma, your graph is wrong, you didn't plot \(y=2x-2\) correctly, and that's where the error comes up. I used your technique with the correct region of integration and got the integrals
$$\int_{0}^{1}\int_{0}^{x}xy\,dy\,dx\,+\int_{1}^{2}\int_{2x-2}^{x}xy\,dy\,dx$$
and ended with the answer \(\frac{5}{6}\) which is the same as what I got from the other order integral
$$\int_{0}^{2}\int_{y}^{\frac{y+2}{y}}xy\,dx\,dy$$
That problem just claimed about 4 hours of my life (Whew)

 

[chisigma]

Observing with care I have noticed that the region is bounded by $\displaystyle y = x,\ y = 2 x - 2,\ y=0$ and not by $\displaystyle y = x,\ y = 2 - 2 x,\ y=0$ so that my previous answer is not correct...

... very sorry! (Crying)...

Kind regards

$\chi$ $\sigma$

 

[Prove It]

Actually Chisigma, your graph is wrong, you didn't plot \(y=2x-2\) correctly, and that's where the error comes up. I used your technique with the correct region of integration and got the integrals
$$\int_{0}^{1}\int_{0}^{x}xy\,dy\,dx\,+\int_{1}^{2}\int_{2x-2}^{x}xy\,dy\,dx$$
and ended with the answer \(\frac{5}{6}\) which is the same as what I got from the other order integral
$$\int_{0}^{2}\int_{y}^{\frac{y+2}{y}}xy\,dx\,dy$$
That problem just claimed about 4 hours of my life (Whew)

Please actually show us your attempt at the integration. There's probably just some very small mistake in amongst it which is giving you the wrong answer...

 

[skate_nerd]

Prove It, my meaning in the last message was that I finally got matching results from the two separate ordered integral set ups. The answer is \(\frac{5}{6}\).