Evaluate $\iint\left(x^2+y^2\right)\,dx\,dy$ throughout the area enclosed by the curves $y=8x$, $x+y=6$, $y=4$ and $y=0$ (10 Marks).
I integrated $$\int_0^4\int_{y/8}^{6-y}\left(x^2+y^2\right)\,dx\,dy$$ and $162.625$ as the answer, which confuses me.
When I calculate the trapezium’s area the normal way, I only get $15$ as the answer.
So the problem is, what is the right answer? Is it that my definition about area is wrong? (I assume that integrating the above will give the trapeziums Area, but is it wrong?)
Am I using the right limits for the integral?
I denote the area by $D$. You have to evaluate
$\int_{D}(x^2+y^2)dxdy$
What you have computed in the "normal way" is $\int_{D}1 dxdy$.
Indeed we have $\int_{D}1 dxdy=15$.
Hence compute $\int_{D}(x^2+y^2)dxdy$.