# Changing order of integration for $\int^2_1 \int^{\sqrt{2x-x^2}}_{2-x}f(x,y)\,dy\,dx$

I need to change integration order of $$\int\limits^2_1 \int\limits^{\sqrt{2x-x^2}}_{2-x}f(x,y)\,dy\,dx.$$

So I found out that:

• The region is bound between $1 \le x \le 2$ and $0 \le y \le 1$
• The upper limit, $y = \sqrt{2x-x^2}$, in terms of $x$ is: $x = 1 + \sqrt{1 - y^2}$
• The lower limit, $y = 2-x$, in terms of $x$ is: $x = 2 - y$

Also, by plugging the value of the limits of $x$ to the limits of $y$, we get: \begin{align} & y_{\,\text{upper}\,}(x=2) = \sqrt{2\cdot 2 - 2^2} = 0 \\ & y_{\,\text{lower}\,}(x=1) = 2 - 1 = 1 \end{align}

Therefore the above integration equals to: $$\int\limits^0_1 \int\limits^{1 + \sqrt{1 - y^2}}_{2 - y}f(x,y)\,dx\,dy \,??$$ Or, perhaps it equals to (changing location of limits of y):

$$\int\limits^1_0 \int\limits^{1 + \sqrt{1 - y^2}}_{2 - y}f(x,y)\,dx\,dy.$$ How do I tell which is correct? (Assuming I was right until now!)
The domain is the patterned area in the following image: • $0\rightarrow 1$ is correct. Since the original integral has limits in positive direction, after change of limits, they should still be in positive direction. You don't need to plug in to get the upper and lower limits of $y$. Just looking at the integration region is enough. – KittyL May 8 '15 at 11:27
• the lower bound for $x$ should be $2+y$. – Math-fun May 8 '15 at 11:46
• @KittyL: Why? We start from $y_1(x=1) = 2 - 1 = 1$ and end at $y_2(x=2) = \sqrt{2\cdot 2 - 2^2} = 0$. Therefore it should be from 1 to 0. – Dor May 8 '15 at 11:53
• @Math-fun: Why? The lower limit is $y = 2 - x$, so $x = 2 - y$. – Dor May 8 '15 at 11:54
• ups, sorry, had a mistake :-) – Math-fun May 8 '15 at 18:47

Your limits for $x$ are correct. Now from the definition of double integral, we integrate the function $f(x,y)$ over the integration region. The result is the volume of the object bounded between $f(x,y)$ and $xy$-plane. It is calculated the similar way as single integral, where we go from the left boundary of $x$ to the right boundary of $x$.
$$\int^1_0\int^{1+\sqrt{1-y^2}}_{2-y}$$
So the terms "upper" and "lower" do not depend on the original order. They should depend on the current situation. In this case, for $y$ values, the upper limit is the upper bound of all values of $y$, the lower limit is the lower bound of all values of $y$.