Please use the best method available to find the volume of t

Solution

actually don\'t think you want the disk method here, as you\'ll see. The method to use is the shell method. But, going as far as we can go with disks/washers: Since the solid is revolved around the y axis, you can\'t really say the radius of the disk is the y value. (A disk with radius y would fall both outside and inside this solid, and there\'s no way of dividing the solid up in this way.) In fact, since the circular disks are centered on the y axis, their radii is the x value, which, if y = sin(x), then x = arcsin(y). However, it\'s a bit more complicated than this, too. Since the circular disks will have \"holes\" in the middle, we use a modified disk method called the \"washer method.\" You will slice the solid into a stack of \"washers,\" find a function for the area of each washer, and then integrate from the lowest bound for y to the highest (in this case, from 0 to 1). Now, from y = 0 to y = sin(pi/4), the solid is a thick, \"regular\" washer, so you can use simple geometry with no integration to calculate its volume. (It\'s the volume of a cylinder of height sin(pi/4) and radius 3pi/4, minus the volume of a cylinder of the same height and radius pi/4. From y = sin(pi/4) to y = 1, it\'s a little trickier. Because of the sinusoidal curve of the upper part of the solid, the outer radius of the washer will be decreasing as the inner radius is increasing. The inner radius r[i] as a function of y can be given by solving y = sin(x) for x, that is, r[i] = x = arcsin(y). However, how do we find the outer radius? x actually has two values on this interval for any particular value of x, but arcsin(y) will only yield the smaller value, the one closest to the y axis. Notice that the function is symmetrical about x = pi/2, that is, both the inner and outer radii of our washer have the same distance from that central point. So, the outer radius r[o] of our washer is arcsin(y) + 2(pi/2 - arcsin(y)). So, the area of our washer will be pi r[o]^2 - pi r[i]^2 = pi(arcsin(y) + 2(pi/2 - arcsin(y))^2 - arcsin^2(y)) Integrate that from y = sin(pi/4) to y = 1, and you\'ve got the volume for the upper portion of your solid.