# Double Integrals

$(a)$ 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!!

• You are asking too many homework questions. It won't particularly benefit you to get any particular question answered because you do not seem to have a strong grasp of the underlying material, so I would suggest that you go to a professor or TA and/or reread your textbook more thoroughly and then think harder about these questions on your own. Constantly tossing out homework questions is not the purpose of this site. – Qiaochu Yuan May 21 '11 at 20:01
• 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. – user4645 May 23 '11 at 10:24

• @user4645: First identify and sketch carefully your region. Then suppose you are integrating first with respect to $y$. Then $y$ goes from "bottom curve" to top curve." So if bottom curve is $y=f(x)$ and top curve is $y=g(x)$, integrate with respect to $y$ from $f(x)$ to $g(x)$. Then integrate with respect to $x$ from the first $x$ to the last. Sometimes, as in (b), top and bottom curves change. Then break up your region into parts where they don't change. In (b), $3$ parts will do the job. You really need someone at a blackboard pointing, or a video. This stuff is very visual. – André Nicolas May 21 '11 at 20:57
1. The integral is taken over the upper semicircle, so $$\int\limits_{-2}^2\:dy\int\limits_0^\sqrt{4-y^2}xe^{(4-x^2)^\frac32}\:dx = \int\limits_0^2xe^{(4-x^2)^\frac32}\:dx\int\limits_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\:dy = 2\int\limits_0^2x\sqrt{4-x^2}e^{(4-x^2)^\frac32}\:dx =$$$$-\dfrac23\int\limits_0^2e^{(4-x^2)^\frac32}\left((4-x^2)^\frac32\right)'\:dx = -\left.e^{(4-x^2)^\frac32}\right|_0^2 = e^8-1$$