`f(x)=xcos(x-sinx)` , `0<=x<=3`
The graph of `f(x)` intersects the `x`-axis when `x=a`, `a!=0`.What is the value of `a`? Also, if the graph of f is revolved 360° about the x-axis from `x=0` to `x=a`. What is the volume of the solid formed?
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To solve this problem, we'll have to be a bit clever. Let's start by finding where the graph intersects zero:
`0 = acos(a-sina)`
Because we're considering values of `a` that are not zero, we can simplify by dividing both sides by `a`:
`0 = cos(a-sina)`
Now, we simply look at possibilities where a cosine is zero: `(2k+1)/2 pi`, where `k in ZZ`.
`(2k+1)/2 pi = a - sina`
Unfortunately, we cannot solve for `a` algebraically at this point. So, we'll need to solve for it numerically for each possible value of `(2k+1)/2 pi`. We'll need to use the Taylor series to get the following result:
`(2k+1)/2 pi = a - (a - a^3/(3!) + a^5/(5!) - ...)`
Eliminating the parenthesis, we end up with the following series result:
`(2k+1)/2 pi = a^3/(3!) - a^5/(5!) + a^7/(7!) - ... = sum_(n=1)^oo (-1)^(n+1)/((2n+1)!) a^(2n+1)`
To approximate a solution, we can just stick with the first 3 terms in the series, which will give us an answer within one hundredth. In other words, we'll solve the following equation numerically:
`(2k+1)/2 pi ~~ a^3/(3!) - a^5/(5!) + a^7/(7!)`
Using a spreadsheet, you can find the following few solutions:
If `k = 0` , a = 2.31
If `k = 1` , a = 3.97
If `k = -1` , `a= -2.31`
Based on these answers and our given domain (`0<=x<=3`), it's clear that the only possibility we must be concerned about is when `k = 0`.
Therefore, our final solution is `a = 2.31`.
Now, in order to find the area of a solid of rotation around the x-axis between `x = 0` and `x = a`, we'll need to solve numerically again. Recall the formula to solve for a solid of revolution by the disc method:
`V = int_0^a pi f^2(x) dx`
In our case, we would need to find the following integral:
`V = pi int_0^2.31 x^2cos^2(x-sinx) dx`
We cannot solve this integral algebraically, so let's proceed numerically. Let's solve for values of `f^2(x)` numerically by Newton's method and then use the trapezoid rule to find the integral.
To use Newton's method, we use the following approximation for `n in NN`:
`f^2(nDeltax) ~~ f_n ^2 = (f_(n-1) + Deltax*d/dx f((n-1)Deltax))^2`
Basically, we are adding an amount equivalent to the rise over run to the function to find the next point. To use this approximation, we must arbitrarily determine a reasonable step size, let's say `Deltax = 0.01` because our `a` is approximated to hundredths. We must also find the derivative of our squared function, so that we may approximate our function values:
`(df(x))/(dx) = d/dx (xcos(x-sinx))`
`= cos(x-sinx) - x(1-cosx)sin(x-sinx)`
Now that we've established how we're finding our function, let's use the following sum which will give us the trapezoid rule approximation of our function's integral. Here, N will be the number of steps, which will be 231 in our case (because `a = 2.31` and `Deltax = 0.01`)
`V = pi int_0^a f^2(x)dx ~~ pi (Deltax)/2(f^2(0) + f^2(a) + sum_(n=1)^(N-1) 2f^2(nDeltax))`
Now, f(0) and f(a) are zero, so we can simplify the above summation:
`V~~pi Deltax sum_(n=1)^(N-1) f^2(nDeltax)`
Putting this information in a spreadsheet using the starting point of f(0) = 0, we will find the following result for our volume:
More terms, smaller steps, and better approximations for `a` will yield a more accurate result. If you use a calculator integrator, you will get 5.89, which is not that far off from our result! However, using that tool is much less instructive!
I hope this helps!
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