Find the area of the surface obtained by rotating the curve y= 1+2 x^2 from x=0 to x = 9 about the y-axis
thats for volume
area of the surface might mean surface area
yea - i assumed thats what caro required - oh i see what you mean!
i cant remember the SA formula @ caro9999 which is it ?
|| 2pi y ds
opps for y rotation, || 2pi x ds
solve limits and inverse so that everything speaks in terms of x=
y= 1+2 x^2 sqrt((y-1)/2) = x when x=0, y=1 x=9, y=163
\[\int_{9}^{163}2pi\ \sqrt{\frac{y-1}{2}}\sqrt{1+(\frac{1}{4\sqrt(\frac{y-1}{2})})^2}\ dy\]
these things are horrible!! where you getting these from?
Calc 2, I know they are the death of me right now lol
\[\int_{9}^{163}\pi\ \sqrt{2y-2}\sqrt{1+\frac{1}{8y-8}}\ dy\] \[\int_{9}^{163}\pi\ \sqrt{\frac{8y-7}{4}}\ dy\] \[\frac{\pi}{16 }\int_{9}^{163}8\sqrt{8y-7}\ dy\] almost got something, may not be right but i can dbl chk it with thte wolf
am I doing this wrong\[S=2\pi\int_{0}^{9}x\sqrt{1+16x^2}\]?this should not be too hard ..or am I having one of those days where I am talking nonsense?
8(8y-7)^(1/2) --> 2(8y-7)^(3/2)/3 2pi(8(136)-7)^(3/2)/48 - 2pi(8(9)-7)^(3/2)/48 = 4583.803... now for the dbl chk :)
this is rotated about the y, not the x
isnt that\[2\pi\int xds\]?
yes, ds is sqrt(1+x'^2)
since we are given y=; we have to convert to to x=
yeah..wait, why do we have to covert it?
so that we have the x and the ds in terms of x
I'm confused... wolf your answer and I'll do mine and we'll see what happens I guess
or rather, to make the y the independant
4583.803 didn't work :/
4583.96 is the wolfs on mine
mine was supposed to be 6114.2 I hope you have multiple tries
or a button that gives it the finger lol
i wonder if this is simpler in polars
2 * pi * y * dx y = 1+2 x^2
it's 2*pi*x*ds that much we agree on
Yup, @TuringTest 6114.2 worked!
oh yeah ... groovy
Thanks for your help guys!!
welcome, I hope you understood the process!
oo .. along the x axis!!
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