Assuming that 4" is the height of the bowl, we can conclude that it is ellipsoidal, like a rugby football, with the shorter axis (x axis) being 4.5" in diameter and its longer axis (y axis) needing to be determined from the given figures. Two domes are cut off each end of the ellipsoid leaving a 3.5" diameter circular top and a 3" diameter circular base. The general equation of the ellipse that will be used to model the bowl is x^2/2.25^2+y^2/a^2 (2.25" being the "radius" of the x axis, and a the radius of the y axis). The ellipse has its centre at the x-y origin. To find a we work out in terms of a where the base of the top dome has a width of 3.5" and where the base of the bottom dome has a width of 3". Putting x=1.75 (half of 3.5) in the equation of the ellipse, we get y=sqrt(a^2(1-1.75^2/2.25^2))=4asqrt(2)/9. Similarly for the base dome: y=asqrt(1-1.5^2/2.25^2)=-asqrt(5)/3, negative because it's below the centre or origin. The difference between these two y values is the height of the bowl, 4", so 4asqrt(2)/9+asqrt(5)/3=4, and a=36/(4sqrt(2)+3sqrt(5))=2.9114 approx.
The volume of the bowl is found by considering thin discs of radius x and thickness dy so that we can use the integral of (pi)x^2dy (the volume of a disc) between the y limits imposed by the heights of the two domes. x^2=5.0625(1-y^2/a^2). Because a is a complicated expression, we'll just use the symbol for it. We can also write 5.0625 as 81/16. The integrand becomes (81(pi)/16)(1-y^2/a^2)dy, with limits y=-asqrt(5)/3 and 4asqrt(2)/9. The integrand is simply the sum of the volumes of the discs between the bases of the two domes.
Integration gives us (81(pi)/16)(y-y^3/(3a^2)) between the limits, which I calculated to be 53.39 cu in approx.
How many candies fit into this volume? We know the length of the candies is 1/2", but we don't know any other dimensions, so we don't know the volume of each candy. Also, the alignment of candies will improve the number of candies fitting into the bowl, but if this cannot be arranged, we have to assume they will be randomly orientated. We could make the assumption that they are cone-shaped with height 0.5" and base diameter 0.25" (radius 0.125" or 1/8") giving them a volume of (1/3)(pi)0.125^2*0.5=0.0082 cu in each approximately. So divide this into 53.39 and we get about 6,526, with very little space between the candies. A more realistic figure can be obtained by finding out how many candies fill a cubic inch when tightly packed. Two will fit lengthwise into an inch. Think of the candies as cuboids 1/4" square and 1/2" long (volume=1/4*1/4*1/2=1/32 cu in). That gives you 32 in a cubic inch, over 1,700 in a bowl. The cuboid is a sort of container that will hold just one candy and allow it some lengthwise freedom of movement.
At the other extreme think of the candies as 1/2" cubes (volume=1/8 cu in) then only 8 will fit into a cubic inch and only about 427 will fit into the bowl. But the candies have greater freedom of movement in a cubic container.
Let's say you can get 100 into one cubic inch packed tightly, then you would get about 5,340 in the bowl. Get a small box and pack in as many candies as you can. Measure the volume of the box (length*width*height). If N is the number of candies, then each has an average effective volume of (box volume)/N. Use this number to divide into the volume of the bowl.
