# Calculating double integrals, getting wrong result

Double integral given:

D is the are between two functions: and

answers given (which I can not get):

I'm getting 0 as the result:

The graph looks like this:

and by looking at the graph I set my solution to be the following (x interval:0-4), y-interval (right function-upper bound, left function-lower bound):

What Am I doing wrong?

• I assume you mean the intersection of the parabolas. The boundaries of integration seem quite off. For instance, $dxdy$ at the end means (to me) that the outer integral is w.r.t. $y$ and the inner is w.r.t. $x$. But then: 1) Why does $y$ range in $[0,4]$, since in the graphic it ranges in $[0,2]$? 2) why do the boundaries of the inner integral depend on the variable $x$, instead of the variable $y$ ? 3) Even allowing that, why are they in the form $ax+b$, since the boundaries are clearly grapics of functions in the form $x=ay^2+by+c$ ? – user228113 Aug 3 '16 at 10:59
• On the other hand, if we assume the outer integral to be wrt $x$ and the inner integral to be wrt $y$, then point $(3)$ stands: the boundaries of integration should be $[0,f(x)]$, where $f(x)$ is piece-wise defined and of the form $\sqrt{b+ax}$. – user228113 Aug 3 '16 at 11:03
• @G.Sassatelli y^2 ranges from 2x to 8-2x and x ranges from 0,4. Why can 't i write the equation like this... – Eugen Sunic Aug 3 '16 at 11:07

I'm assuming you mean the hourglass-like shape between the functions. In that case you should split up the integral into two parts, one part where the first function minors the other and another for the reverse situation. Also in the $y$-bounds you need square roots since the area is given in terms of $y^2$.
• Because the area is defined differently on both intervals. On $[0,2]$, the $y$-range is from $\sqrt{2x}$ to $\sqrt{8-2x}$, while it is the other way around on $[2,4]$. – Sanderr Aug 3 '16 at 11:14