Consider evaluating the double integral
I = ∫ ∫R (2x2 + y)dx dy
where R is the region bounded by y = x and the curve y = x^2. Sketch the region R and confirm by direct calculation that the integral is independent of the order of the integration.
Consider evaluating the double integral
I = ∫ ∫R (2x2 + y)dx dy
where R is the region bounded by y = x and the curve y = x^2. Sketch the region R and confirm by direct calculation that the integral is independent of the order of the integration.
After graphing the region, the limits of integration are pretty obvious. We see that the graphs intersect at $\displaystyle x=0$ and $\displaystyle x=1$. The function $\displaystyle y=x$ is above the graph $\displaystyle y=x^2$ for the particular region described. Thus we can set up the integral:
$\displaystyle
\int_{0}^{1}{\int_{x^2}^{x}{2x^2+y}\,dy}\,dx$
Try evaluating it from here.
Then try it, reversing the order of integration
$\displaystyle
\int_{0}^{1}{\int_{y}^{\sqrt{y}}{2x^2+y}\,dx}\,dy$
You should get the same answer for both. Hope this helped!