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$$\int \int y^2\sin(x^2+y^2)dxdy$$ calculate over D region which is defied as :

$D:{(x,y) x^2+y^2\leq\pi,|x|\leq y}$

(I sketched the region)

we started to study it today and teacher solved in a very peculiar way..

$y=rsin\theta , x=rcos\theta , 0 \leq r \leq \pi , \pi/4 \leq \theta \leq 3\pi/4 $

coverting the integral to polar coordinates we get:

$$\int_{0}^{\sqrt \pi} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}r^2\sin^2\Theta \sin(r^2)r drd\Theta $$ then he did something which I don't fully understand. He wrote the intgral in the following way : $$\int r^3\sin(r^2)dr \int \sin^2\theta d\theta$$ Is it valid to write the integral like this(dividing the integral into 2 inetgrals regarding the fact that there is multiplication in the original integral)?

  • $\begingroup$ As long as your limits are not function of other variable $\endgroup$
    – Rishi
    Sep 6, 2019 at 17:16

3 Answers 3


Yes, it is valid.

If a function $f(x,y)$ can be factored as the product of a function of $x$ and a function of $y$, then we can do that.

Suppose $f(x,y)=g(x)h(y)$ and the double integration is over the domain $R=[a,b]×[c,d]$. From Fubini's Theorem,

$ \iint_Rf(x,y)dA=\int_c^d\int_a^bg(x)h(y)dxdy=\int_c^d\left[\int_a^bg(x)h(y)dx\right]dy $

Observe that in the inner integral, $y$ is a constant, so $h(y)$ is a constant. Therefore, we can write

$ \int_c^d\left[\int_a^bg(x)h(y)dx\right]dy=\int_c^d\left[h(y)\left(\int_a^bg(x)dx\right)\right]dy=\int_a^bg(x)dx\int_c^dh(y)dy $

The last equality is obtained observing $\int_a^bg(x)dx$ is a constant.

Remember, also, that this is generically written for the two dummy variables $x$ and $y$. This is valid for any function of two variables. Particularly in your case, $f(r,\theta)=g(r)h(\theta)$ and the relation still holds.


Yes, it's a valid operation. Basically, you are allowed to do $$ \int_{x_1}^{x_2}\int_{y_1}^{y_2}f(x)g(y)\,dydx = \int_{x_1}^{x_2}f(x)\,dx\cdot \int_{y_1}^{y_2}g(y)\,dy $$ (which is what your lecturer did, only with other names for the variables). It may be more comfortable to do it in two steps, though.

First, if we look only at the inner integral, we are allowed to move $f(x)$ outside, because for all the inner integral cares, that's just a constant factor. So we get $$ \int_{x_1}^{x_2}\int_{y_1}^{y_2}f(x)g(y)\,dydx = \int_{x_1}^{x_2}f(x)\cdot \left(\int_{y_1}^{y_2}g(y)\,dy\right)dx $$ But now, if we look at the outer integral, $\int_{y_1}^{y_2}g(y)\,dy$ is just a constant factor, so we can move that outside. (Note that the integration bounds do not depend on $x$. This is crucial, as otherwise it wouldn't be a constant factor.) This gives us the second step $$ \int_{x_1}^{x_2}f(x)\cdot \left(\int_{y_1}^{y_2}g(y)\,dy\right)dx = \int_{x_1}^{x_2}f(x)\,dx\cdot \int_{y_1}^{y_2}g(y)\,dy $$


Yes since the integrand is product of two functions one of which is only a function of $r$ and the other is only a function of $\theta$ we can split the integrals as your professor did.


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