Volume of Solids of Revolution Hinemann Exercise 6.3 Q13

[macostar]

I worked out part a but the rest is giving me a headache.

[Jacyan]

It's always good to draw these up so you can visualise the rotation.

b) y=r (a horizontal line) is rotated about the x-axis to form a solid of revolution. If you imagine this horizontal line that is "r" units away from the x-axis revolving around the x-axis, you will find that this forms a cylinder. Without restricting the domain of y=r, y=r is a line that goes left and right forever. So when we find the line over the given domain [0,h] we now know that the solid is a cylinder with radius r and a height of h. Hence the volume of V=pi*(r^2)*h.

c) Spin the graph of the circle around the y-axis. This forms a sphere with a radius of r.

d) First we need to find the equation of the line passing through the points (0,3) and (6,5). Use rise over run to find the gradient (m), then use one of the points to find (c) in y=mx+c. Now we can use integration (use the formula) over the domain of this line to find the volume. Give this a go first

Note: I think this is a dodgy question. To form a truncated cone, it should be a line segment joining (0,3) and (6,5) rotated about the x-axis. Without a restriction on the domain, a line rotating about the x-axis will form a solid with infinite volume.

e) This is the same as part d) expect we now have the points (0,r1) and (h,r2). First find the equation of the line joining these 2 points. Remember that h, r1, and r2 are just normal numbers. Now again, we just use the solid of revolution formula to find the volume but this time your answers will in be terms of r1, h and r2. I'll leave the rest up to you but feel free to ask for help if you are still stuck

[macostar]

Hi thanks, for the help. I got all of it except for q13(E) I cant get the answer which is v=pie*h/3 (r2square+r1r2+r1square)