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Find the area between the functions $x+y = 2$ and $x + 4 = y^2$.

The question is relatively simple:

The area between the functions is:

$$\int^{2}_{-3} 2-y-y^2+4 \text{ }dy$$

But can the above area be found by integrating a function with respect to $dx$? As in can the area between the functions be defined as the integration of functions of $x$? (i.e. integration of $f(x)$ and $g(x))$?

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    $\begingroup$ This depends how you express the functions: either $y=2-x$ and $y=|\sqrt{4+x}|$ or $x=2-y$ and $x=y^2-4$ and then you have to change the bounds of integration accordingly $\endgroup$ – Alex Jan 23 '14 at 1:51
  • $\begingroup$ But the second function isn't $y = |\sqrt{5+x}|$, its $y^2 = \sqrt{5+x}$ $\endgroup$ – dfg Jan 23 '14 at 2:06
  • $\begingroup$ @Alex The functions are different because the first one is half a hyperbola, the second is a full one $\endgroup$ – dfg Jan 23 '14 at 2:06
  • $\begingroup$ I think you mean a parabola... $\endgroup$ – colormegone Jan 23 '14 at 2:13
  • $\begingroup$ @dfg: if $y^2=x$ then when you take the square root you get $y=\pm \sqrt{x} = |\sqrt{x}|$ $\endgroup$ – Alex Jan 23 '14 at 2:17
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The graph gives you an idea of what would be involved (and it is often a good idea, for "area between two curves" problems, to have at least a sketch of the geometric region at hand):

enter image description here

You would need to break the calculation into two separate integrals, one between the "upper" and "lower" arms of the parabola over the interval $ \ [ -4, 0] \ $ , the other between the line and the "lower" arm on the interval $ \ [ 0, 5 ] \ . $ So the area would be found from

$$ \int_{-4}^0 \ [ \ \sqrt{x+4} \ ] \ - \ [ \ -\sqrt{x+4} \ ] \ \ dx \ \ + \ \ \int_0^5 \ [ \ 2 - x \ ] \ - \ [ \ -\sqrt{x+4} \ ] \ \ dx \ . $$

In principle, such area integrals can be performed by integration with respect to either coordinate variable. For some problems, it's about the same amount of work either way. Sometimes, though, the geometry of the region makes integration in one direction much simpler, since it may be possible to use a single interval to span the entire integration, but not so in the other variable. In other problems, it may be that the functions for the curves are easier to integrate (or even express) for one variable, but do not have convenient inverses (or ones that can be integrated in closed form) in using the other variable. (So there may be implicit functions for those latter integrations, but that won't help us evaluate the area integral...)

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  • $\begingroup$ Great, thanks a lot! $\endgroup$ – dfg Jan 23 '14 at 16:37

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