# 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 $int sin^2(x)cos(x)dx=1/3 sin^3(x)dx$.