# Volume of Solid of Revolution about Slant Line

We know how to find the volume of a solid of revolution obtained by rotating a region about a horizontal or vertical line (see Method of Disks/Rings and Method of Cylindrical Shells). But what if we rotate about a slanted line, that is, a line that is neither horizontal nor vertical?

Let `C` be the arc of the curve `y=f(x)` between the points `P=(p,f(p))` and `B=(q,f(q))` and let `R` be the region bounded by `C` , by the line `y=mx+b` (which lies entirely below `C`), and by the perpendiculars to the line from `A` and `B`.

We first will try to find area of region `R`. For this we will need another sketch.

We again approximate area by the sum or rectangles, but this time rectangles are perpendicular to the line `y=mx+b`. Area of i-th rectangle is `L_iDelta u`. Thus, approximate area of `R` is `A~~sum_(i=1)^nL_i Delta u`.

Taking `n` very large we have that `A=lim_(n->oo)sum_(i=1)^n L_i Delta u`.

We need to express `L_i` and `Delta u` in terms of `Delta x`.

From the sketch we immediately have that `tan(alpha)=f'(x_i)` (slope of tangent line) and `tan(beta)=m`.

Also, `D_i=(Delta x)/cos(alpha)` , and `Delta u=D_icos(beta-alpha)=D_i(cos(beta)cos(alpha)+sin(alpha)sin(beta))`.

From this we have that `Delta u=(Delta x)/(cos(alpha))(cos(beta)cos(alpha)+sin(alpha)sin(beta))=Delta x (cos(beta)+tan(alpha)sin(beta))`.

Now, we need to express `sin(beta)` and `cos(beta)` in terms of `tan(beta)`.

From identity `tan^2(beta)+1=1/(cos^2(beta))` we have that `cos(beta)=sqrt(1/(1+tan^2(beta)))=sqrt(1/(1+m^2))`.

Since `sin^2(beta)+cos^2(beta)=1` then `sin(beta)=m/sqrt(m^2+1)`.

Therefore, `Delta u=Delta x(1/(sqrt(m^2+1))+f'(x_i) m/sqrt(m^2+1))=Delta x (1+mf'(x_i))/(sqrt(m^2+1))`.

Now, we need to determine `L_i`.

It is fairly easy to do: `L_i=(f(x_i)-(mx_i+b))cos(beta)=(f(x_i)-mx_i-b)/(sqrt(m^2+1))`.

That's all. `A=lim_(n->oo)sum_(i=1)^n L_i Delta u=(f(x_i)-mx_i-b)/sqrt(m^2+1)*(1+mf'(x_i))/(sqrt(m^2+1))Delta x=1/(m^2+1)lim_(n->oo)sum_(i=1)^n(f(x_i)-mx_i-b)(1+mf'(x_i))`.

This is limit of the Riemann sum, i.e. definite integral, so

**Area of region R** is `A=1/(m^2+1)int_p^q (f(x)-mx-b)(1+mf'(x))dx`.

Now, to find volume we use method of disks: we slice perpendicular to the line `y=mx+b`: radius of disk is `L=(f(x)-mx-b)/sqrt(m^2+1)`.

Thus, `A(x)=piL^2=pi((f(x)-mx-b)/(sqrt(m^2+1)))^2`.

Therefore, `V=int_p^q piL^2du=int_p^q pi((f(x)-mx-b)/sqrt(m^2+1))^2(1+mf'(x))/sqrt(m^2+1)dx`.

**Volume of Solid of Revolution around Slant Line **`y=mx+b` is `V=pi/(m^2+1)^(3/2)int_p^q(f(x)-mx-b)^2(1+mf'(x))dx`.

**Example**. Find the area of region `R` bounded by `f(x)=x+sin(x)`, by the line `y=x-2`, and by the perpendiculars to the line from `P=(0,0)` and `Q=(2pi,2pi)`. Find volume of the solid obtained by rotating the region `R` about the line `y=x-2`.

Here we have `m=1` , `b=-2` , `f(x)=x+sin(x)`, thus, `f'(x)=1+cos(x)`.

Therefore, `A=1/(1^2+1)int_0^(2pi)(x+sin(x)-x+2)(1+1*(1+cos(x)))dx=`

`=1/2int_0^(2pi)(2+sin(x))(2+cos(x))dx=`

`=1/2 int_0^(2pi)(4+2cos(x)+2sin(x)+sin(x)cos(x))dx=`

`=1/2int_0^(2pi)(4+2cos(x)+2sin(x)+1/2sin(2x))dx=`

`=1/2(4x+2sin(x)-2cos(x)-1/4cos(2x))|_0^(2pi)=1/2(8pi-2-1/4+2+1/4)=4pi`.

And volume is

`V=pi/(1^2+1)^(3/2)int_0^(2pi)(x+sin(x)-x+2)^2(1+1*(1+cos(x)))dx=pi/sqrt(8)int_0^(2pi)(2+sin(x))^2(2+cos(x))dx=`

`=pi/sqrt(8)int_0^(2pi)(4+4sin(x)+sin^2(x))(2+cos(x))dx=`

`=pi/sqrt(8)int_0^(2pi)(8+4cos(x)+8sin(x)+4sin(x)cos(x)+2sin^2(x)+sin^2(x)cos(x))dx`

Now we use double angle formulas and integral can be rewritten as

`=pi/sqrt(8)int_0^(2pi)(8+4cos(x)+8sin(x)+2sin(2x)+1-cos(2x)+sin^2(x)cos(x))dx=`

`=pi/sqrt(8)int_0^(2pi)(9+4cos(x)+8sin(x)+2sin(2x)-cos(2x)+sin^2(x)cos(x))dx=`

`=pi/sqrt(8)(9x+4sin(x)-8cos(x)-cos(2x)-1/2sin(2x)+1/3sin^3(x))|_0^(2pi)=`

`=pi/sqrt(8)(18pi-8-1+8+1)=pi/sqrt(8)*18pi=(9pi^2)/sqrt(2)` .

Note, how we used substitution `u=sin(x)` for integral `int sin^2(x)cos(x)dx` to obtain that `int sin^2(x)cos(x)dx=1/3 sin^3(x)dx`.