Question

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.)

Answer 1

""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."

Answer 2

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.

Answer 3

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.

Answer 4

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)

Answer 5

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.

Answer 6

...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.

Answer 7

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.

Answer 8

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).

Answer 9

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).

Answer 10

Outer radius squared minus inner radius squared (thickness of the washers is dy): \[\LARGE V = \pi \int\limits_{0}^{3}\left[ 2^2 - \left( 1+ \sqrt[4]{\frac{ y }{ 3 }} \right) ^2 \right] dy\]

Answer 11

"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...

Answer 12

Yeah, it was wrong, my bad. I mixed up my assumptions and what it physically resembled.

Answer 13

Also, wait a minute, the answer should be 12pi/5.