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**Calculus II question…just trying to understand what’s going on….not homework?**

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*Identify the radius and height of each shell and compute the volume.
The region bounded by y=x, y= -x, and x=1 revolved about the y axis.
The solutions manual teslls me that radius=x and height =2x. I am not understanding why this is. I can compute the volume once I figure out how to form the equation but I cannot figure out how they are finding the radius and height.
I am taking an online CALC II course and the system is down for the weekend so I can’t get any help from the instructor.Your help is greatly appreciated. Hopefully someday I can return the favor.
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hey, here goes hope i can help .. take a paper and a pen and draw the lines y=x and y=-x they are basicly the diagonal lines between the x-axis and y-axis that meet at the center .. then u draw the x=1 line which is the line parallel to the y-axis passing through the point (x=1,y=0) so if u look at the region envolved betwen these three lines its basicly a triangle .. and then ur asked to evolve that triangle aroung the y-axis .. the basicly the radius is the distance between the y-axis at the center (x=0,y=0) and the x=1 line which is the side of the triangle … so basicly the distance=radius=x=1 … now the height is the length of the side of the triangle that is farthest from the centre .. again thats on theline x=1 … so to get the thre points of the triangle u can do the intersection between the y=x line and the x=1 line ull get the first point .. and the intersection between the y=-x line and the x=1 line ull get the second point the third one is of course at the center (0,0) … the if u do the distance between the two other points its basicly equal to 2 … to do the volume of the whole thing u need to get the volume of the triangle and then multiply it by 2*Π (two pie that is) cause ur revolving a triangle 360 degrees around an axis .. i tried to explain things as much as possible .. hope i could help

**Calculus III- Graphing a Three Dimensional Plane**

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