Circumference: (of one)
Circumference times length: (of one)
So, now we sum it all up over the interval.
This integral isn't difficult, a simple substitution should do.
from 0 to 1 equals
;-)
Friday, July 24, 2009
Problem #11
Surface area! Oh my, what a wonderful thing. So here is what we are delving into today/tonight. We want to calculate the surface area when the cubic function from 0 to 1 is revolved about the y-axis. The way in which I conceptualize this process is by thinking of a cylinder without a top or bottom and calculating the surface area of it as it gets really small and we add infinite many of them. Thus, we need the circumference and length of one and then integrate.
Circumference: (of one)
 "2\pi (x^3)")
Circumference times length: (of one)
\sqrt{1+9x^4}dx "2\pi (x^3)\sqrt{1+9x^4}dx")
So, now we sum it all up over the interval.
\sqrt{1+9x^4}dx "2\pi \int_{0}^{1}(x^3)\sqrt{1+9x^4}dx")
This integral isn't difficult, a simple substitution should do.
\sqrt{1+9x^4}dx ; u=1+9x^4; du=36x^3dx "2\pi \int_{0}^{1}(x^3)\sqrt{1+9x^4}dx ; u=1+9x^4; du=36x^3dx")
^\frac{3}{2} "\frac{\pi }{27}(1+9x^4)^\frac{3}{2}")
from 0 to 1 equals
}{27} "=\pi \frac{(10\sqrt{10}-1)}{27}")
;-)
Circumference: (of one)
Circumference times length: (of one)
So, now we sum it all up over the interval.
This integral isn't difficult, a simple substitution should do.
from 0 to 1 equals
;-)
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