X=0 is a vertical line, specifically it is the y-axis of the coordinate plane.
6y-5x=0
6y=5x
y=5x/6
and the third line is
3y=21-x
y=(21-x)/3
putting aside the first as it is a vertical line, we need to know when
y=(21-x)/3 is equal to the line y=5x/6 so
5x/6=(21-x)/3
15x/6=21-x
15x=126-6x
21x=126
x=42/7
x=6
so we know that the two sloped lines meet at x=6 or more specifically the point (6,5)
and when x=0 these two lines meet the x=0 line at (0,0) and (0,7)
So you have three points: (0,0),(0,7), and (6,5) forming a triangle.
The base of which is the vertical line from (0.0) to (0,7) which means the base is 7 units in length. The height (in the x direction) is 6 units because the apex is at (6,5)
So the area of a triangle is just bh/2 and in our case is:
A=7*6/2=21 u^2
Now for the volume of this shape rotated about the y-axis....
We will have to do two integrations because of the way the lines are...
First note that we are rotating about the y axis thus x is the radius of revolution so we need our lines expressed in terms of y instead of x...
The top line was y=(21-x)/3...solving for x
x=21-3y and y varies from 5 to 7
The volume is the integral:
V=pSf(y)^2 dy
V=pS441+9y^2-126y dy
V=p[441y+3y^3-63y^2] for y=[5,7]
V=p(1029-1005)
V=24p that is the top half volume...now for the lower half...
