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$(a)$ Sketch the region of integration in the integral

$\int_{y=-2}^{2} \int_{x=0}^{\sqrt{4-y^2}} x e^{{(4-x^{2})}^{3/2}} dx dy$

By changing the order of integration, or otherwise, evaluate the integral.

$(b)$ Let $R$ be the region in the $x-y$ plane defined by $0 \leq x \leq y \leq 2x$, $1 \leq x+2y \leq 4$. Evaluate: $\mathop{\int\int}_{R} \frac{1}{x} dx dy$

I understand how to draw these but I am not sure how to caluculate the limts in either case (especially part $b$).

Can someone explain how we calculate the limits for integration? Once I know that I am sure I can integrate the function myself. Thanks!!

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    This is not a homeowrk question. Im preparing for an exam and understand the steps but Ive never had a quadrilateral area to work out. The only examples we were given were triangles. Therefore I still dont know how to find the limits.2011-05-23

2 Answers 2

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The first thing, as stated, is to sketch the region. Only then calculate the limits of integration. So, what does your sketch look like?

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    sorry i have corrected my mistakes!2011-05-30
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  1. The integral is taken over the upper semicircle, so $\int\limits_{-2}^2\:dy\int\limits_0^\sqrt{4-y^2}xe^{(4-x^2)^\frac32}\:dx = \int\limits_0^2xe^{(4-x^2)^\frac32}\:dx\int\limits_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\:dy = 2\int\limits_0^2x\sqrt{4-x^2}e^{(4-x^2)^\frac32}\:dx =$$ -\dfrac23\int\limits_0^2e^{(4-x^2)^\frac32}\left((4-x^2)^\frac32\right)'\:dx = -\left.e^{(4-x^2)^\frac32}\right|_0^2 = e^8-1$