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Evaluate the double integral $$\iint_D x y \ dy dx$$ Where D is the region bounded by $y^2=x, \space y=1+x^2, \space y=-1, \space x=-1, \space x=1$. The answer given by the lecturer is $\frac{9}{8}$ but I haven't been able to get $\frac{9}{8}$.

Any help would be appreciated. Here is a sketch of D

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    What did you do? Did you make a sketch of the region? That would be very helpful. – 2017-01-19
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    Draw the picture the describes $D$ and get an idea of where to start. You may have integrated incorrectly - without more context, I can't give you details. – 2017-01-19
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    I'm trying to understand what the upwards parabola $\;y=1+x^2\;$ has to do with this...it doesn't even intersects the horizontal parabola $\;x=y^2\;$ ... – 2017-01-19
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    imranfat I've put the image into the question for you – 2017-01-19
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    DonAntonio the parabola is the upper limit for the y values, without it the y values would blow up to infinity and the integral would be divergent – 2017-01-19
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    I think your lecturer was mistaken - I'm fairly confident the answer is 0. I've tried breaking up the region a couple different ways. Logically if you look at the region, you can see there are volumes that will 'cancel out' - like the square in the 3rd quad versus that square reflected over the x axis into the 2nd quad, because of the symmetry of xy and how it changes sign in different quadrants. See here- the red and blue regions cancel, green and orange cancel, and purple and black cancel: https://www.desmos.com/calculator/laym7qk1of – 2017-01-19

2 Answers 2

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Let me decompose your domain into two region, by splitting your domain using the line $y=1$:

$D_1$ is the region that bounded above by $y=x^2+1$, bounded below by $y=1$, bounded from the left by $x=-1$ and bounded on the right by $x=1$.

$$\iint_{D_1} x y \ dy dx=0$$

because for every $(x,y) \in D_1$,$(-x,y) \in D_1$.

$D_2 = D \setminus D_1$.

$D_2$ can also be described as the region that is bounded above by $y=1$, bounded below by $y=-1$, bounded on the left by $x=-1$ and bounded on the right by $y^2=x$.

For this region, for every $(x,y) \in D_2$, $(x,-y) \in D_2$. Hence

$$\iint_{D_2} x y \ dy dx=0$$

Hence

$$\iint_{D} x y \ dy dx=\iint_{D_1} x y \ dy dx+ \iint_{D_2} x y \ dy dx= 0$$

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Just for the sake of seeing if I'm wrong, I'm getting $0$. I broke this problem into four regions (probably inefficient).

$R_1=\int\limits_{-1}^{0}\int\limits_{0}^{1+x^2}xy\,dy\,dx=\frac{-7}{12}$
$R_2=\int\limits_{-1}^{0}\int\limits_{-1}^{0}xy\,dy\,dx=\frac{3}{12}$
$R_3=\int\limits_{0}^{1}\int\limits_{\sqrt{x}}^{1+x^2}xy\,dy\,dx=\frac{5}{12}$
$R_4=\int\limits_{-1}^{0}\int\limits_{-1}^{-\sqrt{x}}xy\,dy\,dx=\frac{-1}{12}$

Those all sum to $0$.

To see how I broke up the regions: http://imgur.com/a/IBMLN