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Calculus: volumes of solids, and substitution!
Loren Michael
Registered User regular
Registered User regular
Okay, so I got this problem:
The curve y = 3x - x^2 between x = 1 and x = 3 is rotated about the x-axis. Estimate the volume of the solid formed using Simpson's rule with four subintervals.
Okay. So. I know how to use Simpson's rule, marginally. And I know how to find volumes. Marginally. But I'm not sure how to combine the two. Do I just go with pi(f(3x - x^2))^2 dx and apply Simpson's rule from there?
Also!
I'm starting to work with substitution, but I'm blanking on some pretty basic parts of the notation, and as such the way of getting from start to finish isn't so clear to me. Here's the example I'm working with:
Find f 2x(x^2 + 3)^.5
Taking some excerpts from the example: "If we assume u to be what's inside the root sign, u = x^2 + 3, the differential of u is du = 2x dx ---> dx = du/2x"
...
So that's saying, the derivative of u is 2x times the derivative of x, and from there, the derivative of x is the derivative of u over 2x? Help me out. Please!
From there, it continues: "Then our integration will be: f 2x(u)^.5 dx = f 2xu^.5 du/2x = f u^.5 du = (u^1.5)/1.5 + C = 2/3*(x^2 + 3)^1.5 + C"
It's that bolded area that I'm not totally getting, but the whole thing is a little unclear. I'm on the verge of getting it, I'm just having a ton of anti-epiphanys that are getting in my way.
The curve y = 3x - x^2 between x = 1 and x = 3 is rotated about the x-axis. Estimate the volume of the solid formed using Simpson's rule with four subintervals.
Okay. So. I know how to use Simpson's rule, marginally. And I know how to find volumes. Marginally. But I'm not sure how to combine the two. Do I just go with pi(f(3x - x^2))^2 dx and apply Simpson's rule from there?
Also!
I'm starting to work with substitution, but I'm blanking on some pretty basic parts of the notation, and as such the way of getting from start to finish isn't so clear to me. Here's the example I'm working with:
Find f 2x(x^2 + 3)^.5
Taking some excerpts from the example: "If we assume u to be what's inside the root sign, u = x^2 + 3, the differential of u is du = 2x dx ---> dx = du/2x"
...
So that's saying, the derivative of u is 2x times the derivative of x, and from there, the derivative of x is the derivative of u over 2x? Help me out. Please!
From there, it continues: "Then our integration will be: f 2x(u)^.5 dx = f 2xu^.5 du/2x = f u^.5 du = (u^1.5)/1.5 + C = 2/3*(x^2 + 3)^1.5 + C"
It's that bolded area that I'm not totally getting, but the whole thing is a little unclear. I'm on the verge of getting it, I'm just having a ton of anti-epiphanys that are getting in my way.
Loren Michael on
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http://curvebank.calstatela.edu/volrev/volrev.htm
Looks like you are on the right track.
I'll do the second bit first so I have a bit of time to write out the first part.
Okay, so you want to integrate
what do you know from calculus, when you derive x^2+3?
well, just do that
d/dx (x^2+3) = d/dx (x^2) + d/dx(3) = 2x + 0 = 2xOkay, so if we were deriving, say, (x^2+3)^4, we also know by the chain rule that:
d/dx (x^2+3)^4 = (4(x^2+3)^3)*d/dx(x^2+3) = (4(x^2+3)^3)*2xokay: what how does this help with the problem? It seems that the chain rule provides us with a way to derive a complicated function by going through the order of operations
now look at the original derivation I did but like this, instead:
d/dx (x^2+3)^1 = (1(x^2+3)^0)*d/dx(x^2+3) = 1*2x (because anything to the power of 0 is 1) = 2xso when you look at the way we find this derivative, it provides a method of 'going backwards' to find an antiderivative. This method won't work for all integrals, and if you keep on learning calculus you will find that there are all kinds of tools available at your disposal.
But think of it like this: The chain rule says:
d/dx f(g(x)) = (d/dx f)(g(x))(d/dx g(x))
in more understandable notation:
d/dx f(g(x)) = f'(g(x))*g'(x)
integrating both of these sides (using INT() as your standard undefined integral):
INT(d/dx f(g(x))) = int (f'g(x))*g'(x))
i'm just gonna switch sides for convenience
INT[(f'g(x))*g'(x)] = INT[(d/dx f(g(x))]
but you know by the fundamental theorem of calculus that when you take the derivative of the integral or the integral of a derivative you get the same function
so
INT[(f'g(x))*g'(x)] = INT[(d/dx f(g(x))]
implies
INT[(f'g(x))*g'(x)] = f(g(x))
so the problem is reduced to finding two functions f and g
alright, now that we have all this motivation, go back to the original problem
h(x) = 2x(x^2+3)^0.5
what is your g'(x)? What is your f'(g(x))?
now start working with your original way of working. If this doesn't make sense by itself I'll run through the computation with you as well. Solution to the first problem coming soon.
2x(x^2+3)^(1/2) dx : with u = x^2 + 3 and du = 2x dx
then you must replace all of you x related terms with the equivalent u terms
2x(u)^(1/2) du/2x
Which, lucky enough, gets the 2x term to cancel out otherwise you would have had to find a way to represent that by modifying your initial u statement of u = x^2+3 to remove the leading 2x from the equation. And once you full substitute everything you can solve normally and then replace u with x^2 + 3 and du with 2x dx.
now, when you look at the region you are trying to rotate, it should look like this:
so when you rotate this thing around the x-axis, you have to think of slicing the volume that you will make into little 'discs' of length dx.
The volume of a disc is
pi*r^2*h
where r is the radius of the circle, h is the height. For us, the radius is the function itself (so 3x-x^2) and the height is the small length dx.
so you just have to integrate pi(3x-x^2)^2
over x between 1 and 3
this is doable for you without estimation, but for simpson's rule you basically estimate an integral via quadratic polynomials
from the wikipedia article:
I assume more terms means dividing the integral into segments of length 0.5, so your original integral from 1 to 3 would be from 1 to 1.5 plus 1.5 to 2 plus 2 to 2.5 plus 2.5 to 3
and then applying simpson's rule to each of those integrals