Calculate the volume of rotation of $C: x = t^2, y = 2t$ and the line $y = x$

Question

I am working on a problem (Pearson Core Pure Mathematics, Book 2, Chapter 4 Mixed Questions, Challenge) and cannot understand where I am going wrong. Below I will post 2 separate methods of mine, where I get the same result, then the official answer. I think my problem is not transforming dx in some way (discussed below).

The question:

Method 1:

Let a and b be the bounds of integration, a = 4 and b = 0. For each $\delta{x}$, the height $h$ of the segments to integrate across will be the length of the line between C and y=x, which has slope perpendicular to x, i.e. $-1$. $y = 2\sqrt{x}$.

Let (a, a) be the value (x, y) of the curve y=x and (x2, y2) be the values of C. Therefore for any point on C, the height of the segment to integreate is:

$\frac{y2 - a}{x2 - a} = -1$

$ a = \frac{2\sqrt{x}+x}{2}$

$ h = \sqrt{(x2-a)^2 + (y2-a)^2} = \sqrt{((x-\frac{2\sqrt{x}+x}{2})^2 + (2\sqrt{x} - \frac{2\sqrt{x}+x}{2}})^2$

$\pi \int_{0}^{4} h^2 dx = \frac{16\pi}{15}$

Method 2:

We can rotate the curve 45 degrees anti-clockwise and then integrate. The rotation matrix:

$$ \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{bmatrix} $$

puts the new y coordinates of C to $\frac{-1}{\sqrt{2}}(x + 2\sqrt{x})$.

$\pi \int_{0}^{4} (\frac{-1}{\sqrt{2}}x + \frac{1}{\sqrt{2}}2\sqrt{x})^2 dx = \frac{16\pi}{15} $.

I think where I have gone wrong is not accounting for any transformation of x (i.e. not changing the bounds or scale of dx). In the answer, the rotated dt is integreated across. However, I thought about this and still cannot see why you need to tranfrom dx - for each dx on the scale you care about, you first transform y then sum. For method 1, you integreate across dx but for each dx the segment is at an angle. Any help in understanding why this is wrong would be much appreciated!

Answer 1

In the end, I was completely wrong as suspected in not accounting for a transformation in x. Going back to the definition of integration as summing areas of height h, width $\delta{x}$ as x goes to zero, it is clear $\delta{x}$ must be adjusted.

E.g. in the image below, it is clear a small step $\delta{x}$ along the $x$ axis is longer by a factor of $\sqrt{2}$. Similarly, the matrix-transformed $\delta{x}$ is longer than $\delta{x}$. Adjusting the answer by $\sqrt{2}$ gives the correct result.

Answer 2

The given line is $x=\tau , y=\tau$. Intersect its perpendicular through $(\tau , \tau )$, i.e. the line $$x+y= 2 \tau$$ with the parabola $4x=y^2$ - take the point of intersection in the 1st quadrant and express its coordinates in terms of $\tau$. Find the square of the distance between that point and $(\tau , \tau )$ as a function of $\tau$. By the disk method, the volume is $\pi$ times the integral of the distance squared, integrated from $\tau =0$ to $\tau =4$.