Let $f\left(x,\:y\right)\:=\:x^2e^{x^2}$ and let $R$ be the triangle bounded by the lines $x=5$, $x=y/2$, and $y=x$ in the $xy$-plane.
Express $\int _RfdA$ in two different ways.
After sketching the region, I got that the first way to write the integral would simply be:
$\int _0^5\int _x^{2x}\:x^2e^{x^2}dydx$
However, I was stuck on how to write it in the other way. With the region, there is no obvious way to write the integral as one in terms of $y$. I feel that this may mean that I need to have the sum of two different double integrals but I am not entirely sure how that would work in this case.
Would I have to subtract the higher $x=(1/2)y$ line from the $x=y$ line or is there some other way?
Any help would be highly appreciated!
Best Answer
You are correct that the other expression has to be written as a sum of two integrals. Sketching the region, we see that we can split the region into an upper and lower section. The lower section is from $y = 0$ to $y = 5$, and the upper is from $y = 5$ to $y = 10$. In this lower section, $x$ will vary from $y/2$ to $y$, and in the upper section, $x$ will vary from $y/2$ to $5$, since here $y > 5$ and the boundary becomes the vertical line $x = 5$. Writing out the integral, we have $$ \int_{0}^{5}\int_{y/2}^{y} x^2 e^{x^2} dx dy + \int_{5}^{10} \int_{y/2}^{5} x^2 e^{x^2} dx dy . $$