Science:Math Exam Resources/Courses/MATH101/April 2008/Question 02 (c)/Solution 1

The first step to solve this problem is to understand what volume we are trying to compute.

The question states that there is a region bounded by the curves y=5 and y=x+4/x. For large values of x the second curve looks like the line y=x, but there is also a vertical asymptote at x=0. So it seems that it will indeed cross the horizontal line y=5 twice (as it dips down before going back up). We can compute where the intersections are by solving:

$x+{\frac {4}{x}}=5$

which is equivalent to

$\displaystyle 0=x^{2}-5x+4=(x-1)(x-4)$

and has solutions x=1 and x=4. Then we want to know how far down the curve goes, so we look for its local minimum. We compute its derivative:

${\frac {d}{dx}}\left(x+{\frac {4}{x}}\right)=1-{\frac {4}{x^{2}}}$

and hence see that it has two critical points, one at x=-2 and one at x=2. We are only interested in the local minimum which is at x=2. This should be enough details to have you be able to obtain the equivalent of the following picture:

This in particular shows that y=5 is bigger than y=x+4/x.

Now, consider a point x between 1 and 4. The distance from x to x=-1 is the radius of the cylindrical shell and is r=x+1. The height of the function at this point is the difference of the top function y=5 to the bottom function y=x+4/x which is 5-x-4/x. When we rotate, we travel a distance of $2\pi r$ (the circumference) and we are multiplying by $h\Delta x$. Thus, we have

${\begin{aligned}V&=\int _{1}^{4}2\pi (x+1)(5-x-4/x)\,dx\\&=2\pi \int _{1}^{4}(5x-x^{2}-4+5-x-4/x)\,dx\\&=2\pi \int _{1}^{4}(-x^{2}+4x+1-4/x)\,dx\\&=2\pi (-x^{3}/3+2x^{2}+x-4\ln |x|){\bigg |}_{1}^{4}\\&=2\pi ((-(4)^{3}/3+2(4)^{2}+4-4\ln |4|)-(-(1)^{3}/3+2(1)^{2}+1-4\ln |1|))\\&=2\pi ((-64/3+32+4-4\ln |4|)-(-1/3+2+1-0))\\&=2\pi (12-4\ln |4|)\\&=2\pi (12-8\ln(2))\\&=8\pi (3-2\ln(2))\\&=24\pi -16\pi \ln(2)\end{aligned}}$

Where each of the last four answers are equivalent and would be accepted as a final answer.