A curve is given in polar coordinates by 2R cos fi [Area]
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Re: A curve is given in polar coordinates by 2R cos fi [Are
It is difficult to read, but if I understand correctly, you are asked to find out the surface area of the rotated polar function, \(\displaystyle r=2Rcos(\phi), \;\ 0\leq \phi\leq \frac{\pi}{2}\)
The polar equation represents a circle with radius R centered at (R,0).
What do you get when you rotate a circle?. A sphere.
What is the surface area of a sphere with radius R?.
Here is a graph of a circle with radius R=1. Thus, it is the graph of \(\displaystyle 2cos(\phi)\)
But, if you must use the integration, the formula for a surface of revolution in polar is
\(\displaystyle 2\pi\int_{a}^{b}rsin(\phi)\sqrt{r^{2}+(\frac{dr}{d\phi})^{2}}d\phi\)
Since \(\displaystyle r=2Rcos(\phi)\), we sub this in for r, along with the derivative and get:
\(\displaystyle 2\pi\int_{0}^{\frac{\pi}{2}}2Rcos(\phi)sin(\phi)\sqrt{(2Rcos(\phi))^{2}+(-2Rsin(\phi))^{2}}d\phi\)
This looks nasty, but whittles down to practically nothing.
Then, upon integrating, gives the surface area of a sphere of radius R.
That whole thing simplifies down to:
\(\displaystyle 8\pi\cdot R^{2}\int_{0}^{\frac{\pi}{2}}sin(\phi)cos(\phi)d\phi\)
It is kind of an overkill, though. All you really need to know is the formula for the surface area of a sphere with radius R.
It is difficult to read, but if I understand correctly, you are asked to find out the surface area of the rotated polar function, \(\displaystyle r=2Rcos(\phi), \;\ 0\leq \phi\leq \frac{\pi}{2}\)
The polar equation represents a circle with radius R centered at (R,0).
What do you get when you rotate a circle?. A sphere.
What is the surface area of a sphere with radius R?.
Here is a graph of a circle with radius R=1. Thus, it is the graph of \(\displaystyle 2cos(\phi)\)
But, if you must use the integration, the formula for a surface of revolution in polar is
\(\displaystyle 2\pi\int_{a}^{b}rsin(\phi)\sqrt{r^{2}+(\frac{dr}{d\phi})^{2}}d\phi\)
Since \(\displaystyle r=2Rcos(\phi)\), we sub this in for r, along with the derivative and get:
\(\displaystyle 2\pi\int_{0}^{\frac{\pi}{2}}2Rcos(\phi)sin(\phi)\sqrt{(2Rcos(\phi))^{2}+(-2Rsin(\phi))^{2}}d\phi\)
This looks nasty, but whittles down to practically nothing.
Then, upon integrating, gives the surface area of a sphere of radius R.
That whole thing simplifies down to:
\(\displaystyle 8\pi\cdot R^{2}\int_{0}^{\frac{\pi}{2}}sin(\phi)cos(\phi)d\phi\)
It is kind of an overkill, though. All you really need to know is the formula for the surface area of a sphere with radius R.
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Re: A curve is given in polar coordinates by 2R cos fi [Are
Thanks a lot man, do you think this problem is a common one to get on an exam in Engineering Calculus 2?
Of course you can't know that, but would you consider it to be a "student kind of problem" or just a problem to understand theory from.
It's the only problem of its kind that I have seen on my problem list.
Thanks a lot man, do you think this problem is a common one to get on an exam in Engineering Calculus 2?
Of course you can't know that, but would you consider it to be a "student kind of problem" or just a problem to understand theory from.
It's the only problem of its kind that I have seen on my problem list.
D
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Re: A curve is given in polar coordinates by 2R cos fi [Are
I have seen this problem - ot it's ilk - many a times on Calculus exam (engineering). This type of calculation is needed when you are considering flow-losses through a pipe (in fluid dynamics).
I have seen this problem - ot it's ilk - many a times on Calculus exam (engineering). This type of calculation is needed when you are considering flow-losses through a pipe (in fluid dynamics).