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My question: $$\iint_D xy \,dA\,,$$ 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 inner integrand represent?

Anyway here's how I set up the double integral with respect to y as the inner integrand:

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

Arctic Char
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Jwan622
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1 Answers1

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Limits $[-1,5] \times[-2,4]$ gives whole rectangle and cannot be true. If you "stay" on axis $OY$ having head up to $OX$ direction, then floor is $y^{2} = 2x + 6$ and roof $y=x-1$.

So you have limits $$\left\{\begin{array}{} -2 \leqslant y \leqslant 4\\ \frac{1}{2}y^2-3 \leqslant x \leqslant y+1 \end{array}\right\}$$ i.e. $$\int\limits_{-2}^{4}\,dy\int\limits_{\frac{1}{2}y^2-3}^{y+1} xy \,dx$$

zkutch
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