I have to evaluate the double integral: $\displaystyle I=\iint_D dxdy$, where $D=\{(x,y)\in\mathbb{R}^2 : -2 \leq y\leq 1 , y^2+4y \leq x \leq 3y+2 \}$ (region of type II).

I've found that $\displaystyle I=\frac{9}{2}$.

Now, I want to write the region $D$ as a type I region, of the form $D=\{(x,y)\in\mathbb{R}^2 : a \leq x\leq b , \phi_1(x) \leq y \leq \phi_2(x) \}$ and evaluate $I$ again.

I've found that $x$ moves in the interval $[-4,5]$ but I can't find the bounds $\phi_1(x)$ and $\phi_2(x)$ of $y$.

Can someone help me?


First we make a little sketch of the region of integration:

Figure 1

Then the bounds on $x$ are set by the smallest and biggest values of $x$ in the whole region. You have it right in that $-4\le x\le5$. Then draw a vertical line through the region (a curve of constant $x$), the red line in the figure. Look at where it enters on the bottom (at $x=3y+2$ or $y=\frac13x-\frac23$) that's the lower bound for $y$. Then where it leaves on the top (at $x=y^2+4y$ or $y=-2+\sqrt{x+4}$ making sure we take the right branch) and that's the upper bound for $y$. The area comes out to be $$\begin{align}A&=\int_{-4}^5\int_{\frac13x-\frac23}^{-2+\sqrt{x+4}}dy\,dx=\int_{-4}^5\left[-2+\sqrt{x+4}-\frac13x+\frac23\right]dx\\ &=\left[-\frac43x+\frac23(x+4)^{\frac32}-\frac16x^2\right]_{-4}^5=-12+18-\frac16(25-16)=\frac92\end{align}$$ In agreement with your previous result.

  • $\begingroup$ Thank you very much for your excellent explanation! $\endgroup$
    – dmvlt
    Jun 3 '16 at 20:02

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