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I want to calculate the following areas of f over S :

a) $S$ is given by the area enclosed by : $y=1, x=3, y=2x$ and $\displaystyle f(x,y) = \frac{x^{2}}{y^{2}}$.
b) $S$ is given by the area enclosed by : $x=y^{2}-4$, $x=3y$ and $f(x,y)= 1$

I calculated the following for a)$$\int_{\frac{1}{2}}^{3}\left(\int_{1}^{2x}\frac{x^{2}}{y^{2}} dy\right) dx.$$

for b) I switched $x$ with $y$ and calculated this: $$\int_{-1}^{4}\left(\int_{-3}^{x^{2}-4} 1dy\right)dx.$$

Is this correct? I draw $S$ and look at the intersection points. Still I have problem determining the right boundaries for my integrals. Can you give me a hint/idea what is a good way to find boundaries?
Merci.

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When you say "area", do you mean volume? –  Dylan Moreland Dec 14 '11 at 18:21

1 Answer 1

up vote 1 down vote accepted

Part a) looks good.

For part b), just draw the given region (without switching $x$ and $y$).

The region in question is bounded by a parabola opening to the right with vertex at $(-4,0)$ and the line passing through the origin of slope $1/3$. The points of intersection are $(-3,-1)$ and $(12,4)$. So, the region is bounded on the left by the parabola and bounded on the right by the line.

You would run into problems integrating with respect to $y$ first, because the functions giving the top and bottom bounds of the "representative vertical line segment at $x$" changes across $x=-4$ to $x=12$.

Things are rosy if you integrate with respect to $x$ first. Here, the inner integral will have limits that correspond to the left and right endpoints of a representative horizontal line segment (the maroon line segment in the diagram below): $$ \int_{x=y^2-4}^{x=3y} 1\,dx. $$

You then integrate the "horizontal line segment integrals" as they range from $y=-1$ to $y=4$: $$ \int_{-1}^4 \int_{ y^2-4}^{3y} \,dx \,dy. $$


enter image description here

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Why integrate in the inner one from $x=4-y^{2}$ to $x=3y$ and not from $y^{2}-4$ to $3y$. And why can you not integrate from $3y$ to $y^{2}-4$ ?? Thanks. –  VVV Dec 14 '11 at 19:17
    
@VVV That was a "typo"; it's fixed now. You have to integrate from the left boundary of the line segment to the right boundary, so it's not from $3y$ to $y^2-4$. –  David Mitra Dec 14 '11 at 19:43
    
I understood this! Thanks. –  VVV Dec 14 '11 at 20:33

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