Integration problem, solids of revolution type (Washer method). Posted before, but i'm still unclear as hell on it. Mind, i'm not trying by any means to be lazy but I seriously need some help on just establishing the limits of integration, inner and outer radius, and some other stuff. Help would be appreciated. (One moment.)
""Find the solume of the solid generated by revolving the region bounded by the x-acis, the curve y = 3x^4, and the lines x = 1 and x = -1 about the line x = 1."
I'm aware that the volume of the portion of the curve from (0,1) rotated around x = 1 will be exactly pi from another problem. What i'm concerned with is applying the washer method to the section of the function from x = -1 to x = 0 being rotated around x = 1.
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First get a picture, so you can measure the radius and such. Since you kinda have that, outer radius is really just 2 - it's measured from the line x=1 to x=-1. inner radius is 1+x. I'll get a nicer pic for you.
Limits will be the limits for the area - x=-1 to x=0 for the portion you want (the limits aren't affected by the axis of rotation being x=1)
They aren't? Okay, let me take a crack at this; the only way I can describe this is that people have made me really, really uncomfortable and unsure about how crossing over into the negative portion of the x-axis affect limits and everything in general to the point that i'm very edgy about doing almost anything when a solid that crosses into the negative x-axis is involved at all. One min.
...or not, I might be going to bed right now, i'll work on this next morning, but thanks for letting me know that the limits aren't affected.
neg x-axis has no effect really, neither does the choice of axis of revolution - the limits are just the limits of the area that you're going to revolve. But in this case, you'll need x in terms of y, and with the function being symmetrical about the y-axis, that could make things a bit weird I guess.
outer radius is just 2 as seen on the diagram. Inner is 1+x, and you just have to solve y = 3x^4 for x. Limits for the blue area are still -1 to 0 (x-limits, that is - you'd plug those into y = 3x^4 to get y limits, which will be 0 to 3).
Actually the symmetry shouldn't make a difference here, the 1+x should take care of it (since y is always positive). Technically it should show 1-x not 1+x on the graph, since x is negative, but since y is always positive, \[\LARGE r= 1+x = 1+ \sqrt[4]{\frac{ y }{ 3 }}\] ie when y=0, the radius is 1, when y=3, the radius is 2 (which makes sense from the graph).
Outer radius squared minus inner radius squared (thickness of the washers is dy): \[\LARGE V = \pi \int\limits_{0}^{3}\left[ 2^2 - \left( 1+ \sqrt[4]{\frac{ y }{ 3 }} \right) ^2 \right] dy\]
"I'm aware that the volume of the portion of the curve from (0,1) rotated around x = 1 will be exactly pi from another problem." hmm, unless i did something wrong, the volume looks like pi/5: http://www.wolframalpha.com/input/?i=pi+*+integral+from+0+to+3%2C+%28%281-%28y%2F3%29%5E0.25%29%5E2%29+dy (in this case the outer radius is 1-x, inner radius is zero) Note that this problem would probably be easier done with the cylindrical shell method than the washer method...
Yeah, it was wrong, my bad. I mixed up my assumptions and what it physically resembled.
Also, wait a minute, the answer should be 12pi/5.
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