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Write the area $D$ as the union of regions. Then, calculate $$\int\int_Rxy\textrm{d}A.$$

First of all I do not get a lot of parameters because they are not defined explicitly (like what is $A$? what is $R$?).

Here is what I did for the first question:

The area $D$ can be written as:

$$D=A_1\cup A_2\cup A_3\cup A_4\cup A_5.$$

Where:

$$A_1=\{(x, y)\in\mathbb{R}^2: x\geq-1\}.$$ $$A_2=\{(x, y)\in\mathbb{R}^2: y\geq-1\}.$$ $$A_3=\{(x, y)\in\mathbb{R}^2: x\leq1\}.$$ $$A_4=\{(x, y)\in\mathbb{R}^2: x\leq y^2\}.$$ $$A_5=\{(x, y)\in\mathbb{R}^2: y\leq1+x^2\}.$$

First, for me I see that $D$ is the intersection of these regions and not the union. Am I wrong?

enter image description here

P.S. This is a homework.

x.y.z...
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4 Answers4

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You should write $$ D =\{(x,y): -1<x<1, -1<y<1+x^2 \} - \{ (x,y): 0<x<1, -\sqrt{x}<y<\sqrt{x} \} $$

mookid
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2

Hint: the area inside each of the parabolic indentations is $$ \int_{-1}^1(1-x^2)\,\mathrm{d}x $$

robjohn
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The easiest way to do this is to write $A$ as $$ A = Q \setminus \bigl( \underbrace{\{(x,y) \in Q \mid y > 1+x^2\}}_{:=B_1} \cup \underbrace{\{(x,y) \in Q \mid x > y^2\}}_{:=B_2} \big) $$ where $Q = [-1,1]\times[-1,2]$ (i.e. a rectangle), $B_1$ is the missing part at the top and $B_2$ the missing part on the right. Note that, per your picture, $B_1$ and $B_2$ are disjoint.

You can further use that the area of $B_1$ is the same as the area under the curve $f(x) = 1 - x^2$ (why?) and that the area of $B_2$ is twice the area under the curve $g(x) = \sqrt{x}$ (again, why?).

fgp
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  • I think it is $[-1, 1]\times[-1, 2]$ not $[-1, 1]^2$ ? – x.y.z... Apr 13 '14 at 17:21
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    @x.y.z... Yup, noticed that right after posting. Fixed. – fgp Apr 13 '14 at 17:22
  • Finally I get $$\int_{-1}^{1}\int_{-1}^{2}xy\mathrm{d}x\mathrm{d}y-\int_{-1}^{1}\int_{1+x^2}^{2}xy\mathrm{d} x\mathrm{d}y-\int_{0}^{1}\int_{-\sqrt{x}}^{\sqrt{x}}xy\mathrm{d} x\mathrm{d}y\quad\text{?}$$ – x.y.z... Apr 13 '14 at 17:43
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It will take a while but to write as unions you need to break things down I.e. for the top left section you can write as $\{-1\leq x\leq 1, 0\leq y\leq 1\}\cup\{-1\leq x\leq 1,1\leq y\leq 1+x^2\}$. And continue this for the other areas.

Btw $R$ refers to the whole region you have written as unions, and A just refers to the fact that you have an area integral.

Ellya
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