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We were working on finding the area between these three functions in class:
$$y=\sqrt{32x}$$ $$y=4-6x$$ $$y=x^2$$ My question is, is it possible to rotate the coordinate plane so that the x axis matches the linear line of y=4-6x? Would doing this make the integration more or less difficult?

The reason I ask is that you can do something "similar" to this when analyzing the forces acting on an object at rest on an inclined plane.

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2 Answers

(i) Yes you can. Transform, Solve, Transform Back can be a powerful technique. The actual phrase was I think popularized by Polya, but the basic idea is omnipresent in mathematics. And here you don't even have to transform back, since the transformation can be chosen so that it keeps area invariant.

(ii) Anything you do to improve $y=4-6x$ will disimprove the others. The equations of our two parabolas are currently very nice. So two uglifications versus one improvement. Plus the cost of transforming. A cost/benefit analysis argues against doing it for this problem.

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The actual area between the functions would stay constant, would it not, if the planes were rotated and the functions were modified accordingly? Would it be possible to leave the functions as is, find the magnitude of rotation (such as by solving for the triangle made between the linear function and a specified origin, and then deriving an angle) and use this somehow with the remaining two functions? Edit It appears you addressed this comment at least in part. None-the-less, I shall leave it here should you wish to readdress it. –  Evan Jul 19 '12 at 20:07
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Rotating is probably more difficulty that necessary.

Draw the three functions. Figure out where they intersect, so you can figure out the integration bounds. Then integrate the difference of the relevant functions over the relevant interval.

enter image description here

Just for giggles, try a coordinate transformation that aligns the linear function with the x-axis. Use the linear transformation $u = x$, $v=6x+y$. This is not quite a rotation, but is not as messy. The Jacobian of this transformation is $1$, so the area is preserved. It is easy to see that $x = u$, $y = v-6u$, which gives the equations: $$ v=u^2+6u, \ v = 4,\ v = \sqrt{32 u} + 6u.$$ Plotting these equations gives:

enter image description here

I suspect it is a little more work to deal with these equations, but de gustibus non est disputandum.

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To emphasize, generally the first step should be to try and draw a picture. –  copper.hat Jul 19 '12 at 19:04
So with that transformation, you would only need to do one integration with respect to y, instead of two integrations before the transformation? –  Evan Jul 19 '12 at 22:38
Yes, you could, but I don't think there are huge savings... –  copper.hat Jul 19 '12 at 23:23
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