0
$\begingroup$

Double integral given: enter image description here

D is the are between two functions: enter image description here and enter image description here

answers given (which I can not get): enter image description here

I'm getting 0 as the result:

The graph looks like this: enter image description here

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):

enter image description here

What Am I doing wrong?

$\endgroup$
3
  • $\begingroup$ 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 $x=ay^2+by+c$ ? $\endgroup$ – user228113 Aug 3 '16 at 10:59
  • $\begingroup$ 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 $\sqrt{b+ax}$. $\endgroup$ – user228113 Aug 3 '16 at 11:03
  • $\begingroup$ @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... $\endgroup$ – Eugen Sunic Aug 3 '16 at 11:07
1
$\begingroup$

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 $y^2$.

$\endgroup$
3
  • $\begingroup$ Ok I see, but why do I have to split it up? $\endgroup$ – Eugen Sunic Aug 3 '16 at 11:08
  • $\begingroup$ 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 $[2,4]$. $\endgroup$ – Sanderr Aug 3 '16 at 11:14
  • $\begingroup$ I've got i all wrong tnx, I was looking ath the surface below instead of the functions... $\endgroup$ – Eugen Sunic Aug 3 '16 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.