Asked by Suzie
I am in Calculus and am currently learning how to find the Area of a Surface of Revolution. I cannot understand what the surface of revolution (whether it's the x-axis, y-axis, or y=6) is. For example, I had a problem saying to use the washer method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations y=x y=3 x=0 about the line y=4. I don't understand what I am supposed to do. Sometimes I can get the problem right with the way my teacher explained it and sometimes I can't. Please help!
Answers
Answered by
drwls
First of all, you have to visualize what the surface of revolution is. You get it by rotating a line (straight or curved) 360 degrees about an axis, like the outside of a vase or pot being made on a lathe or potter's wheel.
Draw yourself a figure with the three lines plotted. In your case, the line that you are rotating is the straight horizontal line y=3 from x=0 to 3, and the 45 degree line y=x from x = 3 to 4.
When rotating the y = x line and y = 3 line about y=4, you get a pencil-shaped volume running from x=0 to x=4. You will have to do the integration in two parts: from x=0 to x=3, you get a cylinder with radius 1 and length 3 (from x=0 to x=3). The volume of that part is 3 pi. You don't even have to perform the integration to see that. The "washer method" consists of doing a dx integration of pi r^2 dx from 0 to 3, with r = 4-3 = 1. The region from x=3 to 4 is conical because the y=x curve forms the surface. The volume of that part is (1/3) pi
I get the complete volume to be
V = (10/3) pi.
Draw yourself a figure with the three lines plotted. In your case, the line that you are rotating is the straight horizontal line y=3 from x=0 to 3, and the 45 degree line y=x from x = 3 to 4.
When rotating the y = x line and y = 3 line about y=4, you get a pencil-shaped volume running from x=0 to x=4. You will have to do the integration in two parts: from x=0 to x=3, you get a cylinder with radius 1 and length 3 (from x=0 to x=3). The volume of that part is 3 pi. You don't even have to perform the integration to see that. The "washer method" consists of doing a dx integration of pi r^2 dx from 0 to 3, with r = 4-3 = 1. The region from x=3 to 4 is conical because the y=x curve forms the surface. The volume of that part is (1/3) pi
I get the complete volume to be
V = (10/3) pi.
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