Asked by Ryan
                I having a hard time on solving ths question is there any way I can get help on it 
So for a plastic bowl it can be seen as the function:
f(x) = x^12
and from x= 0 to x= 1 in conjunction of the graph:
g(x) = -x^20 + 2
with x = 0.9 to x = 1 and using these to move about the y axis
- how can we use integration to determine the volume?
-how can we utilize cylindrical shell intergration to determine the volume?
-Is there a surface integral that can we can use to find the surface area?
if you can help thanks!
            
        So for a plastic bowl it can be seen as the function:
f(x) = x^12
and from x= 0 to x= 1 in conjunction of the graph:
g(x) = -x^20 + 2
with x = 0.9 to x = 1 and using these to move about the y axis
- how can we use integration to determine the volume?
-how can we utilize cylindrical shell intergration to determine the volume?
-Is there a surface integral that can we can use to find the surface area?
if you can help thanks!
Answers
                    Answered by
            oobleck
            
    the two curves intersect at (1,1) and (-1,1)
Since both curves are symmetric about the y-axis, the volume can be calculated integrating on the interval [0,1], using shells of thickness dx
v = ∫[0,1] 2πrh dx
where r = x and h = (2-x^2) - x^12
The surface is a bit trickier, since the two parts have to be done separately. The bottom would be
∫[0,1] 2πr ds
where r = x and ds = √(1+(12x^11)^2) dx
    
Since both curves are symmetric about the y-axis, the volume can be calculated integrating on the interval [0,1], using shells of thickness dx
v = ∫[0,1] 2πrh dx
where r = x and h = (2-x^2) - x^12
The surface is a bit trickier, since the two parts have to be done separately. The bottom would be
∫[0,1] 2πr ds
where r = x and ds = √(1+(12x^11)^2) dx
                    Answered by
            Ryan 
            
    Thank you, so how I go about determining the cross sections?
    
                    Answered by
            oobleck
            
    Using shells, there are no cross-sections. That's the disc method, which doesn't work too well here, because you cannot express x as a function of y.
And besides, for a solid of revolution, <u>all</u> cross-sections are circles (or washers).
    
And besides, for a solid of revolution, <u>all</u> cross-sections are circles (or washers).
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