$$ (x-y=1), (y=x^2-4x+3), y=3$$
A) I understand rotating about y=3 means it is a horizontal shift, and it is just like rotating about the x-axis except at y=3. Therefore, we will need to write, $$x-y=1$$ in terms of x, which comes out out to: $$y=x-1$$
B) I know the points of intersection between these 2 functions are at x=1, and x=4. so when writing the volume:
$$ V = \int_1^4 π(outer-radius)^2-(inner-radius)^2 dx $$
C) Because we have to rotate about y=3, we find the radius by subtracting the functions from 3 and determine which function is the outer and innder radius with respect to the line y=3.
The inner function would be y=x-1
$$ 3-(x-1) $$ which gives, $$ y=4-x $$
and the outer function $$ 3-(x^2-4x+3) $$ gives $$ 4x-x^2$$
So our volume then becomes,
$$V = π\int_1^4 (4x-x^2)^2-(4-x)^2 dx $$
Is this correct? After integrating I got 735π/5 which can be simplified with 147π. However, the answer key says 108π/5... I must have made an error somewhere.
I like to do this sort of integral by integration by parts via the tabular method. The first column is a polynomial factor, differentiated down to a constant, the second is alternating $\pm1$, and the third is the other factor which can be easily integrated as many times as necessary. $$\begin{array}{c|c|c}&&(4-x)^2\\ x^2-1&(+1)&-\frac13(4-x)^3\\ 2x&(-1)&\frac1{12}(4-x)^4\\ 2&(+1)&-\frac1{60}(4-x)^5\end{array}$$ Then just multiply across the rows and add results to get $$\begin{align}\pi\int_1^4\left[(4x-x^2)^2-(4-x)^2\right]dx&=\pi\int_1^4(x^2-1)(4-x)^2dx\\ &=\pi\left[-\frac13(x^2-1)(4-x)^3-\frac16x(4-x)^4-\frac1{30}(4-x)^5\right]_1^4\\ &=\pi\left[\frac16(1)(3)^4+\frac1{30}(3)^5\right]=\frac{108\pi}{5}\end{align}$$ This approach is easier than it looks because that factor of $(4-x)$ always makes the upper bound expression vanish and the factor of $(x^2-1)$ also makes the lower bound expression vanish in the first term.