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$$\int \int xy dA$$ where D is the region bounded by $y = x - 1$ and parabola $y^{2} = 2x + 6$

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Quick question, what does the integrand represent? I see that 5,4 is an intersection point, but 5 * 4 = 20 and I'm not sure what the integrand is even supposed to represent.

Anyway here's how I set up the double integral:

$$\int_{-1}^{5} \int_{x-1}^{2x+6} xy \,dy \,dx$$ $$\int_{-1}^{5} \int_{x-1}^{2x+6} (y+1)y \,dy \,dx$$ $$\int_{-1}^{5} \int_{x-1}^{2x+6} (y^2 + y \,dy \,dx$$ $$\int_{-1}^{5} \left[(\frac{y^3}{3} + \frac{y^2}{2})\right]_{-2}^{4} \,dy \,dx$$

Jwan622
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2 Answers2

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You did not setup the integration limits correctly... The integration domain is not bounded by below by a single curve. You could for instance use $$ \int_{-2}^4 \int_{y^2/2-3}^{y+1} xy \,dx dy = \int_{-2}^4 y\left[\frac 12 x^2\right]_{\frac 12 y^2-3}^{y+1} dy= \cdots = 36. $$

Regarding the "meaning" of the integral, if you want to think in terms of physical quantities, it could represent the total mass of a thin plate with the shape of the integration domain, if the density was $\rho(x,y)= xy$.

PierreCarre
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  • If I'm not wrong the first integral seems between $-2$ and $4$ and the second one starts from $\frac12 y^2-3$. – user Oct 12 '20 at 13:03
  • @user you are correct, read 16 when it was in fact "+6". – PierreCarre Oct 12 '20 at 13:04
  • But why can't we integrate using dy first? The right boundary is y = x - 1 and the left boundary is the parabola so can't we integrate with respect to dy first? – Jwan622 Oct 12 '20 at 13:12
  • @Jwan622 We just need to look at the region and decide if we are setting the first the bounds for $x$ or "y". If we start with $y$, the integration domain is described as $${(x,y)\in\mathbb{R}^2: -2 \leq y \leq 4, \frac 12 y^2 -3 \leq x \leq y+1}.$$

    The left boundary is represented by $x = \frac 12 y^2 -3$ and the right boundary by $x = y+1$.

    If we start by saying that $-3 \leq x \leq 5$, we cannot place $y$ between two single functions of $x$ and must divide into two regions.

     – PierreCarre Oct 12 '20 at 13:17
  • Think like this: If you integrate first with respect to $y$, it is as if you are colouring the region with vertical lines. If you start from left to right, initially your lines are between two branches of the parabola but, after some point ($x=-1$) those vertical lines are between the straight line (below) and the parabola (above). In order for the integration to work, this cannot happen. – PierreCarre Oct 12 '20 at 13:23
  • So I'm not sure what you mean? In my double intetgral, I'm integrating with respect to y first so I'm looking at the right curve - left curve in my integrand. Is that not right? – Jwan622 Oct 13 '20 at 03:10
  • @Jwan622 when you are integrating with respect to $y$, the integration limits are relative to the functions (of $x$) that bound the domain below and above, not right and left. – PierreCarre Oct 13 '20 at 09:07
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We need to divide the integral as follows

$$\int_{-3}^{-1} \int_{-\sqrt{2x+6}}^{\sqrt{2x+6}} xy \,dy\,dx + \int_{-1}^{5} \int_{x-1}^{\sqrt{2x+6}} xy \,dy\,dx $$

or as an alternative

$$\int_{-2}^{4} \int_{\frac12y^2-3}^{y+1} xy \,dx\,dy$$

user
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  • I find your notation when placing $dx, dy$ misleading. Strictly speaking, $$\int_{-2}^4 y dy \int_{\frac 12 y^2-3}^{y+1} x dx = \left( \int_{-2}^4 y dy \right) \int_{\frac 12 y^2-3}^{y+1} x dx$$ – PierreCarre Oct 12 '20 at 13:13
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    @PierreCarre It is a kind of notation I often use but if it is confusing I change that. Thanks – user Oct 12 '20 at 13:28