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This is prime. It cannot be factored. If you graph this you will see the graph never crosses the x-axis. to see this you might try using quadratic formula
youre an idiot, you can factor out x
5x( x^2 + 2x + 3)
haha very true. sorry i was looking at factoring the remaining part after the x had been factored out
ok , :)
are you good at surface area integral? 2pi y dS
somewhat i haven't used that too much
hmmm, i have a question posted on the left . its about a torus
Surface area of a torus which has radius r which is generated by the circle centered at (R,0) with radius r. And assume R>r. So the equation of the circle is (x-R)^2 +y^2 = r^2. FOr some reason my integral does not work. i have integral 2pi y dS, where y = sqrt [r^2 - (x-R)^2], and ds = sqrt ( 1 + dy/dx ^2). dy/dx = -2(x-R)/ (2sqrt(r^2 - (x-R)^2)). anyways, integration limits , the lower limit is R-r, and upper R+r
ok ill take a look
the answer i should get is 4pi^2 r*R
my integral is int sqrt [r^2 - (x-R)^2] sqrt [ 1 + [2(x-R)/ (2sqrt(r^2 - (x-R)^2))]^2 .
the setup looks correct as do the functions
it prob is a computational error, thats a pretty big integral.
yes it is
is that last part sqrt all squared so it cancels out
hmm nevermind i thought i could reduce it for you. as far as the integration you'll have to find someone else to help you or use some graphing software. sorry
we can reduce it