It is a good question about disk method Can you help me to solve

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Disk method question

Use the disk method to find the volume of the solid obtained by rotating the region bounded by y=x^2 and y=x about the line y=x.
 
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HallsofIvy

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Re: It is a good question about disk method Can you help me to solve:)

From a point (x0, x02) on y= x2, find the line through that point perpendicular to y= x: y= -(x- x0)+ x02. Find where that line crosses y= x and calculate the distance from (x0, x02) to the point of intersection, as a function of x0. That will be the radius of the disk. Since the length of the segment y= x for y from x0 to x0+ [itex]\Delta x[/itex] is [itex]\sqrt{2}\Delta x[/itex] , in place of dx you will want to use [itex]\sqrt{2}dx[/itex] as the "thickness" of the disk when you integrate.

Alternatively, you can rotate the graph through -45 degrees to make the line y= x rotate to y= 0. Since, in this case, x= x' cos 45- y'sin 45, y= x' sin 45+ y' cos 45 or [itex]x= \frac{\sqrt{2}}{2}x'+ \frac{\sqrt{2}}{2}y'[/itex], [itex]y= \frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y'[/itex]. Replacing x and y in y= x with those gives [itex]y= \frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y'= \frac{\sqrt{2}}{2}x'+ \frac{\sqrt{2}}{2}y'[/itex] which reduces to [itex]\sqrt{2}y'= 0[/itex] or y'= 0. Replacing x and y in [itex]y= x^2[/itex] gives the equation of that curve in the x'y' coordinate system and then rotate around the x' axis.
 
Re: It is a good question about disk method Can you help me to solve:)

From a point (x0, x02) on y= x2, find the line through that point perpendicular to y= x: y= -(x- x0)+ x02. Find where that line crosses y= x and calculate the distance from (x0, x02) to the point of intersection, as a function of x0. That will be the radius of the disk. Since the length of the segment y= x for y from x0 to x0+ [itex]\Delta x[/itex] is [itex]\sqrt{2}\Delta x[/itex] , in place of dx you will want to use [itex]\sqrt{2}dx[/itex] as the "thickness" of the disk when you integrate.

Alternatively, you can rotate the graph through -45 degrees to make the line y= x rotate to y= 0. Since, in this case, x= x' cos 45- y'sin 45, y= x' sin 45+ y' cos 45 or [itex]x= \frac{\sqrt{2}}{2}x'+ \frac{\sqrt{2}}{2}y'[/itex], [itex]y= \frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y'[/itex]. Replacing x and y in y= x with those gives [itex]y= \frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y'= \frac{\sqrt{2}}{2}x'+ \frac{\sqrt{2}}{2}y'[/itex] which reduces to [itex]\sqrt{2}y'= 0[/itex] or y'= 0. Replacing x and y in [itex]y= x^2[/itex] gives the equation of that curve in the x'y' coordinate system and then rotate around the x' axis.

I cannot understand that. Can you write integral of first solution, please. Thank you.
 

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