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another one huh :)
Yep.. I'm going to start tutoring next week. I am having the worst time with these..
try drawing out the region to get the bounds
yep, i've got that
so the bounds should be in y? so 0 to 6
I was thinking of using the cylindrical shell method
2pi integral (radius)height
radius = y height = y-36?
i'm not as familiar with the shell method, mostly because i don't like using it anyway, wouldn't the radius be in terms of x, and the height be 6-y
so we would use the disc method?
so radius = y^2 and height is 6-y?
you could use either i believe for this one. if you used the disc method: outer radius = 36 inner radius = 36-y^2 ...yes
so we have the integral of pi(36-36+y^2)?
think area of circle: A = pi*r^2
forgot the ^2
what should you integrate from, or what are the bounds
0 to 36?
the outer radius is 36 right, that is in x-direction now imagine the figure being sliced horizontally so we integrate over y what are the bounds of y
oh 0 to 6 in terms of y
ok so from 0 to 6 integral (36 - 36 + x^2)^2?
not quite its R^2 - r^2 Not (R-r)^2 -> 36^2 - (36-y^2)^2
because you're not subtracting the radius, you are subtracting the areas
ohh.. do I have two separate integrals?
no just keep them separate...you can simplify it
nvm.. i see now
I got it right!! yay.. haha.. =3628.8pi. Thank you so much!
Now it's time to study biology... fun :/