# Transformed integral problem

I've come across a question where it required an integral, $$I=\int\int_Tdpdq$$Where $0\le p \le 1$, $0\le q \le 1$, is the region $T$ to be mapped onto the region R with transformation:$$p(x)=4x-4x^2, q=y$$ Where $0\le x \le 1$, $0\le y \le 1$ for the region R.

I transformed the integral to get:$$I = \int\int_R(4-8x)dxdy$$ Solving both of these intregrals over their given domains I managed to get:$$I=\int\int_Tdpdq=1$$ and$$I = \int\int_R(4-8x)dxdy=0$$ My question is, why are the two values different? Shouldn't the value be the same after the transformation? I may have made an error somewhere or is there some explanation for the results obtained? Any help is appreciated.

Thanks

• Your second integral cannot be over the same region, as you have changed coordinates. As you did not give the specific limits, it is hard to pinpoint the mistake. Note also that there is a trick here: $p(0) = p(1)$! May 4, 2017 at 9:13
• The limits were given as the (p,q) and (x,y) domains from the given question, as such, I set up the integrals as $\int_0^1\int_0^1 dpdq$ and $\int_0^1\int_0^1 (4-8x) dxdy$ May 4, 2017 at 9:22
• I have edited the problem statement to reflect the true nature of the question May 4, 2017 at 10:02
• The image of $T=\{(p,q)\in\mathbb R^2|0\le p\le 1;0\le q\le1\}$ is not $\{(x,y)\in\mathbb R^2|0\le x\le 1;0\le y\le1\}$ bu $\{(x,y)\in\mathbb R^2|0\le x\le 1/2;0\le y\le1\}$ May 4, 2017 at 10:58
• So this would account for the discrepancy in my answers May 4, 2017 at 11:09