Question

let ( f(x,y)=x^{2}e^{x^{2}} ) and let ( r ) be the triangle bounded by the lines ( x = 3 ), ( x=y/3 ), and ( y=x ) in the ( xy )-plane.
express ( iint_{r}f(x,y)da ) as a double integral in two different ways by filling in the values for the integrals below. (for one of these it will be necessary to write the double integral as a sum of two integrals, as indicated; for the other, it can be written as a single integral.)
(a) with one double integral:
( iint_{r}f(x,y)da=int_{a}^{b}int_{c}^{d}f(x,y)d )
where ( a= ), ( b= ), ( c= ), ( d= )

Explanation:

Step1: Find the intersection points

The line \(x = y/3\) can be rewritten as \(y = 3x\). The line \(y=x\) and \(x = 3\).
Inter - section of \(y=x\) and \(x = 3\) gives the point \((3,3)\).
Inter - section of \(y = 3x\) and \(x = 3\) gives the point \((3,9)\).
Inter - section of \(y=x\) and \(y = 3x\) gives the point \((0,0)\).

Step2: Set up the double - integral

We want to express \(\iint_{R}f(x,y)dA\) as \(\int_{a}^{b}\int_{c}^{d}f(x,y)dy dx\).
For the region \(R\) bounded by \(x = 3\), \(x=y/3\) (or \(y = 3x\)) and \(y=x\), when we integrate with respect to \(y\) first (inner integral) and \(x\) second (outer integral):
The limits for \(x\) go from \(x = 0\) to \(x = 3\) (so \(a = 0\), \(b=3\)).
For a fixed \(x\) value, the lower limit for \(y\) is \(y=x\) and the upper limit for \(y\) is \(y = 3x\) (so \(c=x\), \(d = 3x\)).

Answer:

\(a = 0\), \(b = 3\), \(c=x\), \(d = 3x\)