area of Surface of revolution: cone w/ r=6, h=3, about y-axis

[just a girl]

Can anyone help me with this question please:

Find the area of the curved surface of a right-circular cone of radius [FONT=MathJax_Main]6[/FONT] and height [FONT=MathJax_Main]3[/FONT] by rotating the straight line segment from (0,0) to [FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main])[/FONT]about the y-axis.

Many thanks!

 

[mathAssistant]

Washer method

The line has the equation y = mx + b where b=0 because the line passes through the origin. m = (3-0)/(6-0) = 1/2, so the line has the equation y = (1/2)x

Now imagine the cone is formed by rotating that line around the y-axis. You can slice the resulting cone starting from the origin heading up the y-axis (to a height of 6) into washers. Each washer has surface area equal to the circumference, 2(pi)r, times the thin height dy.

Assuming the cone is open at the top:
Area = Integral(0,6)_of ( 2(pi)r*dy )

What is r? r is nothing more than x.
Can x be written in terms of y? Hint: look at the equation of the line. Once you have the area of each washer in terms of y, you can evaluate the definite integral to get the area.

If the cone is considered open that will be your answer, if it is considered "capped" at the top don't forget to add the surface area of the top: (pi)r^2 where r = 3.