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

[erogol]

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.

 

[HallsofIvy]

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+ $\Delta x$ is $\sqrt{2}\Delta x$ , in place of dx you will want to use $\sqrt{2}dx$ 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 $x= \frac{\sqrt{2}}{2}x'+ \frac{\sqrt{2}}{2}y'$, $y= \frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y'$. Replacing x and y in y= x with those gives $y= \frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y'= \frac{\sqrt{2}}{2}x'+ \frac{\sqrt{2}}{2}y'$ which reduces to $\sqrt{2}y'= 0$ or y'= 0. Replacing x and y in $y= x^2$ gives the equation of that curve in the x'y' coordinate system and then rotate around the x' axis.

 

[jack81]

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+ $\Delta x$ is $\sqrt{2}\Delta x$ , in place of dx you will want to use $\sqrt{2}dx$ 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 $x= \frac{\sqrt{2}}{2}x'+ \frac{\sqrt{2}}{2}y'$, $y= \frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y'$. Replacing x and y in y= x with those gives $y= \frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y'= \frac{\sqrt{2}}{2}x'+ \frac{\sqrt{2}}{2}y'$ which reduces to $\sqrt{2}y'= 0$ or y'= 0. Replacing x and y in $y= x^2$ 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.