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The two dimensional integral $$\int_0^{b}dy_1\int_0^{b}dy_2=\int_0^bdy_i=b^2,$$ but when I change variables to $a_i=\sin(\pi y_i/b)$, where $i=1,2$, the integral is zero. There should be a way to turn it to a double integral which is twice the first integral. E.g., $$\int_0^bdy_i=\int_0^1da_i \left|\displaystyle\prod_{i=1}^{2}\frac{1}{\sqrt{1-a_i^2}} \right|+\int_1^0da_i\left|\displaystyle\prod_{i=1}^{2}\frac{1}{\sqrt{1-a_i^2}} \right|=2\int_0^1da_i\left|\displaystyle\prod_{i=1}^{2}\frac{1}{\sqrt{1-a_i^2}} \right|.$$ I think it may have something to do with the determinant of the Jacobian...

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  • $\begingroup$ I suspect you cannot make that substitution since the domains of $y_i$ and $a_i$ differ. $\endgroup$ – Bobson Dugnutt Aug 31 '16 at 18:23
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Let's try the analogue in one dimension. We know that $\int\limits_{0}^{b}{\text{ d}y} = b$. Now let $a = \sin(\pi y/b)$, so that $\text{d}a = \frac{\pi}{b}\cos(\pi y/b)\text{ d}y$.

But wait! You don't have a $\cos(\pi y/b)$ in the integrand! No worries, we'll just introduce it there. That is, we can write $$\int\limits_{0}^{b}{\text{ d}y} = \int\limits_{0}^{b}{\frac{b}{\pi\cos(\pi y/b)}\left(\frac{\pi}{b}\cos\left(\frac{\pi y}{b}\right)\text{ d}y\right)} = \int\limits_{0}^{b}{\frac{b}{\pi\cos(\pi y/b)}\text{ d}(\sin(\pi y/b))}. $$ So as long as we can write $\frac{b}{\pi\cos(\pi y/b)}$ as a function of $\sin(\pi y/b)$, i.e. as long as we can find $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $\frac{b}{\pi\cos(\pi y/b)} = f(\sin(\pi y/b))$, we can apply the substitution rule for integrals to evaluate our integral.

So can we find such an $f$? Well, no. If we could, then that means the value of $\frac{b}{\pi\cos(\pi y/b)}$ depends only on the value of $\sin(\pi y/b)$. But if we look at $y_1 = \frac{b}{4}$ and $y_2 = \frac{3b}{4}$, then $$\sin\left(\frac{\pi y_1}{b}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} = \sin\left(\frac{3\pi}{4}\right) = \sin\left(\frac{\pi y_2}{b}\right)$$ but $$\cos\left(\frac{\pi y_1}{b}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \ne -\frac{\sqrt{2}}{2}= \cos\left(\frac{3\pi}{4}\right) = \cos\left(\frac{\pi y_2}{b}\right).$$ So the value of $\frac{b}{\pi\cos(\pi y/b)}$ can't depend on the value of $\sin(\pi y/b)$ alone. Thus, we cannot apply the substitution rule here.

A similar line of reasoning applies for the two dimensional integral in your question.

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  • $\begingroup$ I think I've got it. With the one dimensional integral $\int_0^ady$ can be written as $\int_0^{a/2}dy+\int_{a/2}^{a}dy=\int_0^{a/2}dy+(-1)\int_{a}^{a/2}dy$, then changing to $a$ co-ordinates we have the integrals $\int_0^{1}|J|da+(-1)\int_{0}^{1}|J|da=0$. But for 2d we have $\int_0^ady_1\int_0^ady_2$, which again splitting the integral and changing variables we have $\int_0^{1}\int_0^1|J|da_1da_2+(-1)^2\int_0^{1}\int_0^1|J|da_1da_2=2\int_0^{1}\int_0^1|J|da_1da_2$. So it seems that when the dimension is even, you can add them together? I don't know if my reasoning is correct? $\endgroup$ – Lewis Proctor Sep 1 '16 at 12:34
  • $\begingroup$ No, your reasoning is not correct. How does $\int\limits_{a}^{a/2}{dy} = \int\limits_{0}^{1}{|J|da}$? By that reasoning, it would imply $\int\limits_{0}^{a}{dy} = 0$--you obviously don't agree with that, right? @LewisProctor $\endgroup$ – Joey Zou Sep 1 '16 at 15:08
  • $\begingroup$ Yeah I agree with you there, but there is a proposition in multivariable calculus which (in the books notation) reads: $\int_{[a,b]\times[c,d]}f(x,y)d(x,y)=\int_{[b,a]\times[d,c]}f(x,y)d(x,y)$. $\endgroup$ – Lewis Proctor Sep 2 '16 at 11:59

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