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:

[![enter image description here](https://i.stack.imgur.com/xD1pfm.jpg)](https://i.stack.imgur.com/xD1pfm.jpg)

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!

[![enter image description here](https://i.stack.imgur.com/aAn1pm.jpg)](https://i.stack.imgur.com/aAn1pm.jpg)