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let $$D=\{(x,y)|y\ge x^3,y\le 1,x\ge -1\}$$

Find the integral $$I=\dfrac{1}{2}\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$$

My idea: $$I=\int_{0}^{1}dx\int_{x^3}^{1}(x^2y+2y^2)dy+\int_{-1}^{0}dx\int_{0}^{-x^3}(xy^2+2x+2y^2)dy$$ so $$I=\int_{0}^{1}[\dfrac{1}{2}x^2y^2+\dfrac{2}{3}y^3]|_{x^3}^{1}dx+\int_{-1}^{0}[\dfrac{1}{3}xy^3+2xy+\dfrac{2}{3}y^3]|_{0}^{-x^3}dx$$

$$I=\int_{0}^{1}[\dfrac{1}{2}x^2+\dfrac{2}{3}-\dfrac{1}{2}x^8-\dfrac{2}{3}x^9]dx+\int_{-1}^{0}[-\dfrac{1}{3}x^{10}-2x^4-\dfrac{2}{3}x^9]dx$$ so $$I=\dfrac{5}{6}-\dfrac{1}{18}-\dfrac{2}{30}+\dfrac{1}{33}+\dfrac{1}{10}-\dfrac{2}{30}=\dfrac{67}{90}$$

My question: my reslut is true? can you someone can use computer find it value?

because I use Tom methods to find this reslut is $$\dfrac{79}{270}$$ which is true? so someone can use maple help me?Thank you

math110
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    Isn't it just $$I=\int\limits_{-1}^1\int\limits_{x^3}^1 (x^2y+xy^2+2x+2y^2)dydx$$ ? Any particular reason you're trying to split the integral ? – AgentS Sep 19 '14 at 15:46

3 Answers3

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It seems that you already realize that the for $D$, $D = \{x^3 \leq y \leq 1\} \cap \{-1 \leq x\}$ is the same as $D = \{x^3 \leq y \leq 1\} \cap \{-1 \leq x \leq 1\}$. So, for a function $f(x,y)$, you should have $$ \iint_D f(x,y) \, dxdy = \int_{-1}^1 \int_{x^3}^1 f(x,y)\,dxdy. $$ However, you seem to have split up your function $f(x,y)$ over different bounds without reason. You really should be integrating $$ \int_{-1}^1 \int_{x^3}^1 (x^2 y + xy^2 + 2x + 2y^2)\,dxdy = \int_{-1}^1 \left[ \frac{1}{2}x^2y^2 + \frac{1}{3} xy^3 + 2xy +\frac{2}{3} y^3 \right]_{y=x^3}^{y=1}\,dx= ... $$

Tom
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The domain $D$ looks roughly like a right triangle $ABC$ with the right angle $B$ at $(-1,1)$, $A$ at $(-1,-1)$, $C$ at $(1,1)$ and a curve $y=x^3$ instead of a straight line from $A$ to $C$. Since the curve does not do anything tricky (one value of $x$ maps to one value of $y$ and vice-versa) you can do this as a single integral, integrating either $x$ or $y$ first.

Your tactical choice of breaking the integral into two regions separated by the Y axis led you to make a simple error in the limits in the second region, which should still have been $\int_{x^3}^{1}$. You also left out some of the terms in the integrand.

The correct answer is $$ I = \int_{x=-1}^{1} dx \int_{y=x^3}^{1}\left( x^2y + xy^2 + 2x + 2y^2 \right) dy $$ $$ I = \int_{x=-1}^{1} dx \left[ \frac{1}{2}y^2 x^2 + \frac{1}{3} y^3 x + 2yx + \frac{2}{3} y^3 \right]_{x^3}^1 $$ $$ I = \int_{x=-1}^{1} dx \left[ \frac{1}{2}x^2 + \frac{1}{3} x + 2x + \frac{2}{3} - \frac{1}{2}x^8 - \frac{1}{3} x^{10} - 2x^4 - \frac{2}{3}x^9 \right] $$ $$ I = 2 \left[ \frac{1}{6} + 0 + 0 + \frac{2}{3} - \frac{1}{18} - \frac{1}{33} - \frac{2}{5} - 0 \right] $$ (here, we use the fact that the integral from $-1$ to $1$ of an odd power of $x$ is zero to drop terms with $x$ and $x^9$) $$ I = \frac{344}{495} $$

Mark Fischler
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$\frac{67}{90}$ doesn't look correct. Here is what wolfram computes

rrr
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