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…

y=5x/6, x=6y/5, and y varies from 0 to 5

V=pS(6y/5)^2 dy

V=(36p/25)Sy^2 dy

V=(36p/25)(y^3/3)

V=(36p/25)(125/3)

V=60p

So the total volume is 60p+24p=84p

≈264 u^3