Let be a nonnegative, differentiable function whose graph passes through both points and . For every real number , the ordinate set of on the interval generates a solid of revolution when rotated about the -axis whose volume is given by

Find the formula for the function .

**Incomplete.**

And I forgot to include the little command at the top. I’ll get this blogging thing… eventually. Here’s my post, hopefully with equations this time!

So, a little change of substituting y = f(x) gave the perfect visual cue to figure it out. Let’s take it from the top

The mystery function in question is such that its ordinate set over [0, x], when rotated about the x-axis (or in our case, the t-axis), sweeps out a solid of rotation with a volume equal to x^2 f(x). Written otherwise:

If we take the derivative with respect to x of both sides, we get

Rearranging terms gives us the derivative of the function in terms of x and the function f(x)

Now, this is where the visual cue comes into play. Let y = f(x), the above equation becomes

Now, if we recall from section 8.25, a first-order homogeneous differential equation f(x, y) is such that

So, we can test our above equation y’ to see if it is homogeneous.

As such, y’ = f(x, y) is a homogeneous first order equation. Now, let y = vx, and y’ = v’x + v, our equation becomes

Rearranging terms and using Leibniz notation gives us the separable equation:

Factoring the left side of the equation gives us:

Which we can change into partial fractions

Solving for A and B gives us

So, our separable equation is now:

Integrating both sides

And taking the exponential of each side

Substituting back v = y/x

Rearranging terms gives us y as a function of x and L

And finally, we can use our initial condition f(1) = 2/pi to solve for L and give y as a function of x

This gives us

Which matches the answer in the back of the book.

Eureka!

So, a little change of substituting y = f(x) gave the perfect visual cue to figure it out. Let’s take it from the top

The mystery function in question is such that its ordinate set over [0, x], when rotated about the x-axis (or in our case, the t-axis), sweeps out a solid of rotation with a volume equal to x^2 f(x). Written otherwise:

\begin{align*}

\pi \int_{0}^{x} [f(t)]^2 dt = x^2 f(x)

\end{align*}

If we take the derivative with respect to x of both sides, we get

\begin{align*}

\pi [f(x)]^2 = 2x f(x) + x^2 f'(x)

\end{align*}

Rearranging terms gives us the derivative of the function in terms of x and the function f(x)

\begin{align*}

f'(x) = \frac{\pi [f(x)]^2 – 2xf(x)}{x^2}

\end{align*}

Now, this is where the visual cue comes into play. Let y = f(x), the above equation becomes

\begin{align*}

y’ = \frac{\pi y^2 – 2xy}{x^2} \\

= f(x, y)

\end{align*}

Now, if we recall from section 8.25, a first-order homogeneous differential equation f(x, y) is such that

\begin{align*}

f(tx, ty) = f(x, y)

\end{align*}

So, we can test our above equation to see if it is homogeneous. Let y’ = f(x, y)

\begin{align*}

f(tx, ty) = \frac{\pi (ty)^2 – 2(tx)(ty)}{(tx)^2} \\

= \frac{t^2(\pi y^2 – 2xy){t^2 x^2}

= \frac{t^2}{t^2} \frac{\pi y^2 – 2xy}{x^2}

= \frac{\pi y^2 – 2xy}{x^2}

= f(x, y)

\end{align*}

As such, y’ = f(x, y) is a homogeneous first order equation. Now, let y = vx, and y’ = v’x + v, our equation becomes

\begin{align*}

v’x + v = \frac{\pi (vx)^2 – 2x(vx)}{x^2} \\

= \frac{x^2}{x^2} \pi v^2 – 2v \\

= \pi v^2 – 2v

\end{align*}

Rearranging terms and using Leibniz notation gives us the separable equation:

\begin{align*}

\frac{dv}{dx}x = \pi v^2 – 3v

\end{align*}

\begin{align*}

\frac{dv}{\pi v^2 – 3v} = \frac{dx}{x}

\end{align*}

Factoring the left side of the equation gives us:

\begin{align*}

\frac{dv}{v(\pi v – 3)}

\end{align*}

Which we can change into partial fractions

\begin{align*}

\frac{dv}{v(\pi v – 3)} = \frac{A}{v} + \frac{B}{\pi v – 3}

\end{align*}

Solving for A and B gives us

\begin{align*}

A = -\frac{1}{3}; \quad B = \frac{\pi}{3}

\end{align*}

So, our separable equation is now:

\begin{align*}

\frac{1}{3} (\frac{-1}{v} + \frac{\pi}{\pi v + 3}) dv = \frac{dx}{x}

\end{align*}

Integrating both sides

\begin{align*}

\frac{1}{3} log|\pi – \frac{3}{v}| = log|x| + C

\end{align*}

And taking the exponential of each side

\begin{align*}

(\pi – \frac{3}{v})^{1/3} = Kx \quad (where \quad K = e^C)

\end{align*}

\begin{align*}

\pi – \frac{3}{v} = Lx^3 \quad (where \quad L = K^3 = e^{3C})

\end{align*}

Substituting back v = y/x

\begin{align*}

\pi – \frac{3x}{y} = Lx^3

\end{align*}

Rearranging terms gives us y as a function of x and L

\begin{align*}

y = \frac{3x}{\pi – Lx^3}

\end{align*}

And finally, we can use our initial condition f(1) = 2/pi to solve for L and give y as a function of x

\begin{align*}

\frac{2}{\pi} = \frac{3}{\pi – L} \\

2\pi – 2L = 3\pi \\

L = \frac{-\pi}{2} \\

\end{align*}

This gives us

\begin{align*}

y = \frac{3x}{\pi + \frac{\pi}{2} x^3} \\

& = \frac{6}{pi} \frac{x}{2 + x^3}

\end{align*}

Which matches the answer in the back of the book.

This one has been driving me nuts… The setup is obvious, but getting from the initial equation to the function f(x) escapes me…

The setup is as such

And if we differentiate both sides with respect to x, we get

Then, if we divide both sides by \begin \pi[f(x)] \end , we get

And the rest, well, I haven’t been able to pull apart. Maybe soon…