The discussion revolves around finding the volume of a solid of revolution formed by rotating the area bounded by the curve x=1+(y-2)² and the line x=2 about the x-axis. Participants are exploring the correct setup for the integral to calculate this volume.
The discussion is ongoing, with participants providing guidance on correcting the integral setup. There are multiple interpretations being explored regarding the heights of the rectangles and the correct limits for integration. Some participants have suggested adjustments to the integral, while others are still grappling with the implications of their setups.
Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is an emphasis on understanding the geometric interpretation of the problem.
Here's where you've gone wrong, I think. The representative rectangles need to be inside the parabola, not outside, as you have set the integral up.ThiagoG said:Homework Statement
x=1+(y-2)^2, x=2. Rotating about the x-axis
Homework Equations
Volume=(2∏y)(1+(y-2)2(Δy)
Limits of integration would be from 1 to 3
2∏∫(y)(1+(y-2)2)dy
eumyang said:Here's where you've gone wrong, I think. The representative rectangles need to be inside the parabola, not outside, as you have set the integral up.
eumyang said:(To help visualize what you did earlier, I attached two pics. The red region is what is being rotated around the x-axis. The "Wrong.bmp" file shows what you did, and the "Right.bmp" file shows what the problem is asking.)
The way you had set it up, your heights of the representative rectangle (or the distance from the y-axis to the parabola) is x (which equals 1 + (y - 2)2). Given that the distance from the y-axis to the line x=2 is 2, can you tell me the height of a rectangle from the parabola to the x=2 line?
eumyang said:No, you don't want to add 2 and 1 + (y - 2)2.
eumyang said:Almost. Switch the order of subtraction.