hi
yrtherenonames,
the best thing is to draw a graph for yourself and then use either the Shells method or the Washer method...
the quiddity of both methods, however, are the same - find the circular area of an slice or the volume of a shell and then sum all such volumes in the specified limits.
Part A (i)
Shell Method:
remember that when using the Shell method, the slices or shells are seated
parallel to the axis about which the area is to be rotated. (ie. in this case, it's the y-axis)
Shells are simply vertical strips taken from the area and rotated about the approrpiate (vertical) axis...
so in your case, imagine you are taking any vertical strip of infinitesimally small width (ie. dx) from the area below y=x^2 and to the left of x =1.
so the dimensions of this strip is then: 'y' (which is the height of the strip) times 'dx' which is the width of the strip... ie. Area =
y*dx
when this strip is rotated about the y-axis it forms a 'shell' of a cylinder (ie. a hollow cylinder). now you want to find the volume of this shell of cylinder, so the best way is to imagine that you are now opening up this shell and laying it on a flat surface - making a (very) thin
rectangular prism.
we know already that two of the dimensions of this prism are 'y' and 'dx' (above). we just need to find the third.
the third dimension is of course the length of the prism - which is the same as the circular
circumference of the base or top of the shell cylinder before you opened it up. so the length is then:
2pi*x ; where 'x' acts as the radius.
hence, the volume of any particular cylindrical shell is then:
dv = 2pi*x*y*dx
so then sum of all shells in the specified region will then give you the total volume that you seek:
ie.
V = 2pi*Int[xydx] from x=1 to x=0; where y = x^2 in this case
so, V = 2pi*Int[x^3 dx] = pi/2*[x^4] , x=1, x=0 --->
V = pi/2 units^3
and that's how you would approach the problem using the method of Shells... just keep practising at it and you'll get it soon enough - remember to 1) imagine the strips first, then 2) opening it up to a flat rect. prism, 3) finding volume of one shell, and then 4) suming the volumes of all the shells up {ie. integrating}.
Washer Method:
the Washer method is similar to the Shell method but with one exception - the initial strips are taken
perpendicular to the axis about which the area is to be rotated. (ie. in this case, it's horizontal strips perpendicular to the y-axis.)
pick a point on the curve y=x^2 and imagine yourself taking a horizontal strip from that point to the y-axis, now for your problem this strip is
NOT a part of the area/region specified.
the dimensions of this strip is length: x , and width/depth: dy
now the outlying limit of the specified area is x =1 ; so any horizontal strip taken from x =1 to the y-axis will have dimensions length: 1 , and width/depth: dy
the Washer is similar to a hollow
disc that is formed when you rotate a horizontal strip of infinitesimally small depth WITHIN the specified area around the y-axis (or any vertical axes);
so since the horizontal strip that you've taken is (as said above) NOT a part of the area, then the volume of this strip after it's been rotated must be subtracted from the volume of a rotated strip from the outlying limit x =1.
after rotation, you need to find the third dimension, just like you did for the Shell.
in this case it's the area of the surface of the disc or washer.
the
inner disc (the you will subtract) has a circular surface area of
A = pi*x^2; hence the volume of this disc is
dv = pi*x^2*dy
similarly, the volume of the disc formed from rotating the horizontal strip of the outlying limit x=1 is:
dv = pi*1^2*dy =
pi*dy
hence, the real volume of the segmented strip within the specified area after rotation is: pi*dy - pi*x^2*dy =
dv = pi*(1 -x^2)dy
therefore, the total volume of the solid of revolution is given by:
V = pi*Int[(1 -x^2)dy] from y =1 to y =0 ; where x^2 =y in this case
so, V = pi*Int[(1 -y)dy] = pi*[y -(1/2)y^2], y=1, y=0 --->
V = pi/2 units^3
{Notice how both the Washer and the Shell method obtains the same volume - which makes sense.}
so that's how you'd approach the problem using the Washer Method... remember to 1) find a horizontal strip within the specified area, 2) rotate that strip around the specified vertical axis (doesn't have to be the y-axis), then 3) find the volume of the washer so formed, and if need be, subtract the volume of an inner disc from the volume of the outer (outlying) disc to find the volume, and 4) suming the volumes of all such washers up between the specified limits {ie. integrating}.
but finally 5) keep practising it until you get the hang of it.
i hope this explanation helps you out a bit
yrtherenonames 
; you should try these methods out on the rest of your questions yourself for practice.
if you get stuck on any of them, then don't hesistate to post again :uhhuh:
Edit: it is crucial to remember however, that under an exam situation, when you come across a question that asks you to find the volume but one that does
NOT say it's necessary to use either the Washer or Shell methods (ie. "hence
or otherwise"), then it would be an advantage to you if you could simply find the volume using 2 unit maths techniques - which you can for the particular problem in this post.
this can potentially save you a lot of time.
