Using Trapezoid Rule Twice huh?

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Homework Statement


trape.png

DSC08494.jpg


Homework Equations


A=(1/2)w[E+2M]
Where w=width between each interval
E=f(a0)+f(an) is the sum of the end values
M=f(a1)+f(a2)+...f(an-1)


The Attempt at a Solution



I found the area of the first trapezoidal rule but can't identify the other
w=(15/5)=3metres
E=2.6+2.4=5
M=3+3.2+2.9+2.4=11.5

Therefore, A=(1/2)w[E+2M]=42

I couldn't identify the 2nd one
 
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In order to calculate the volume of water in the pit, you can calculate the area of each of the 6 cross sections using the trapezoidal rule. Make a table or plot of the resulting cross sectional areas and use the trapezoidal rule again in the direction perpendicular to the cross sections to calculate the volume of water in the pit.
 
You are estimating a volume, so you are doing a 2-dimensional integral of the form [tex]\int \int f(x,y) \, dx \, dy [\tex] where f(x,y) = depth at point (x,y). For each x, you do the trapezoidal rule to estimate [itex]\int f(x,y)\, dy[/itex], and you are doing that at the x-points 0, 3, 6, 9, 12 (meters). Of course, at different x you have different numbers of y-points, so rather than doing the trapezoidal rule twice, I would rather say you are doing it 5 times (once for each of the 5 x-values).<br /> <br /> RGV[/tex]
 
Shouldn't this be in Calculus and Beyond?