find the center of mass of the solid of uniform density biunded by the graphs of the equations
x^2+y^2=a^2 is a cylinder with its circular cross-section in the x-y plane; the cylinder is sliced at an angle by the plane z=cy, forming a wedge. y=0 is the x-z plane and z=0 is the x-y plane; these planes and the conditions that y and z are positive impose further constraints on the shape. The cylinder is sliced in half by the x-z plane so the cross-section will be a semicircle and only the upper half is required. The semicircular end rests on the z=0 plane, and doesn't extend beyond the vertex of the wedge. The length of the wedge is ac, because the slope of the angle of the wedge is tan^-1(c) to the y axis and the radius of the wedge is a. Viewed from the side, along the x axis, the wedge is a right-angled triangle with sides a (height), ac (length) and hypotenuse a√(1+c^2). In everyday terms, the wedge looks like the end of a lipstick that has been sliced across.
Going back to the side view of the wedge, we can see that the flat end has a height a. As we move along the wedge towards the point the height changes and shrinks to zero when we reach the vertex. The length at a point P(y,z) is a-y. This height is the distance of a chord to the circumference, starting at height a. The first task is to work out the area of a segment. We do this by subtracting the area of a triangle from the area of a sector, given the distance of the chord from the circumference. This distance is a-y, so the distance from the centre of the circle is, conveniently, y. y/a=cosø where ø is half the angle subtended by the arc of the sector. So ø=cos^-1(y/a) and the area of the sector is given by 2ø/2π=ø/π=A/πa^2, or A=a^2ø (ø is measured in radians). The area of the isosceles triangle formed by joining the radial ends of the arc of the sector is B=aysinø=y√(a^2-y^2), and the area of the segment is A-B=a^2cos^-1(y/a)-y√(a^2-y^2). (Note that when y=0, this expression becomes πa^2/2, the area of the semicircle.) The volume of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz where infinitesimal dz is the thickness of the segment. The mass of the segment is directly proportional to its volume because the density is uniform, so the volume is sufficient to represent the mass.
The centre of gravity (COG) of the wedge can be found by summing moments about a point G(x,y,z) which is the COG. This sum has to be zero. Since a semicircle is symmetrical in the x-y plane about the y axis, we know the x coord of G must be zero. We need G(0,g,h) where g is the height along the z axis of the COG and h the z position.
We need to find the COG of a segment first. The length of the chord of a segment is 2√(a^2-y^2). Earlier we discovered this when we were finding out the areas of the sector and isosceles triangle. Consider just two dimensions. The length of the chord is one dimension and if we create a rectangle from this length and an infinitesimal width dy, the area will be 2√(a^2-y^2)dy. If we think of the rectangle as having mass, its mass is proportional to its area, so area is a sufficient measure for mass. Let's call the COG of the segment point C(x,y)=(0,q), then the moment of the rectangle about the COG=C(0,q) is mass times distance from C=2(q-y)√(a^2-y^2)dy. If we sum the rectangles over the interval y to a we have 2∫((q-y)√(a^2-y^2)dy)=0 for y≤y≤a. The interval looks strange, but it means that y is just a general value, which is determined when we return to the wedge. What we are trying to do is to find q relative to y.
We need to evaluate the definite integral, so we split it: 2q∫(√(a^2-y^2)dy)-2∫(y√(a^2-y^2)dy. Call these (a) and (b).
To evaluate (a), let y=asinu, then dy=acosudu; √(a^2-y^2)=acosu and the integral becomes 2a^2q∫(cos^2(u)du). Since cos(2u)=2cos^2(u)-1, cos^2(u)=½(cos(2u)+1), so we have a^2q∫((cos(2u)+1)du)=a^2q(sin(2u)/2+u).
Since sin(2u)=2sinucosu, this becomes a^2q(sinucosu+u)=a^2q(y/a * √(1-(y/a)^2)+sin^-1(y/a))=qy√(a^2-y^2)+a^2qsin^-1(y/a). Now we apply the limits for y: πa^2q/2-qy√(a^2-y^2)+a^2qsin^-1(y/a).
To evaluate (b), let u=a^2-y^2, then du=-2ydy and the integral is +∫√udu = (2/3)u^(3/2)=(2/3)(a^2-y^2)^(3/2). Applying the limits we have: -(2/3)(a^2-y^2)^(3/2).
Multiply through by 6: 3πa^2q-6qy√(a^2-y^2)+6a^2qsin^-1(y/a)-4(a^2-y^2)^(3/2)=0.
From this q=4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a)). Fearsome though this looks, when y=0 and z=0, q=4a^3/(3πa^2)=4a/(3π), which is in fact the COG of a semicircle. When y=a and z=ac, q=0. Strictly speaking, we should write q(y) instead of just q, showing q to be a dependent variable rather than a constant.
To find the COG of the wedge, we need to find the moments of the COGs of the segments. For a segment distance y above and cy along the z axis, we have the above expression for q. So q(y) is the y coord of the COG of the segment and cy is the z coord. The x coord is 0. The mass of the segment is (a^2cos^-1(y/a)-y√(a^2-y^2))dz, as we saw earlier, and this mass can be considered to be concentrated or focussed at the COG of the segment (0,q(y),cy). The COG of the wedge is at G(0,g,h) so the distance of the mass of the segment from G is g-q(y) and h-cy in the y and z directions. The y-moment is (g-4(a^2-y^2)^(3/2)/(3πa^2-6y√(a^2-y^2)+6a^2sin^-1(y/a))(a^2cos^-1(y/a)-y√(a^2-y^2))dz and the z-moment is (h-cy)(a^2cos^-1(y/a)-y√(a^2-y^2))dz. The sums of these individual components are zero. These lead to rather complicated integrals, requiring considerably more space than is available to work out, even if I knew how to do so! Read More: ...