Volume Ratios

This take home quiz will revolve around volumes and the cylinder that it is taken from (also, did you notice the pun?)

1. Let the function \(f(t) = at\) where \(a\) is a positive real number. If we revolve the area under the curve from \(0 \leq t \leq x\), where x is a positive real number, around the x-axis, it will create a familiar shape. Draw a sketch of this situation on a set of axes. What is the volume of this solid, in terms of \(x\)?

   

   \[ \begin{align} V(x) &= \pi \int_0^x f^2(t) dt\\ &= \pi \int_0^x (at)^2 dt\\ &= \pi a^2 \int_0^x t^2 dt\\ &= \pi a^2 \left[\frac{t^3}{3}\right]_0^x\\ &= \frac{\pi}{3}a^2x^3 \end{align} \]

2. Now consider the cylinder that is created when revolving the region under the constant function \(y=f(x)\) (where \(f\) is the function from before) in the first quadrant around the x-axis. Draw this cylinder on the same set of axes as before. What is the ratio of the volume of the familiar shape to the cylinder? (Hint: You should recognize that you just proved a formula from geometry).

   

   \[ \begin{align} C(x) &= \pi x \int_0^x f^2(x)dt\\ &= \pi x \cdot (ax)^2\\ &= \pi a^2 x^3\\\\ \frac{V(x)}{C(x)} &= \frac{\frac{\pi}{3}a^2x^3}{\pi a^2 x^3}\\ &= \frac{1}{3} \end{align} \]

3. Now let \(g(t)=at^3\). Find the ratio of the solid formed when revolving the region under \(g(t)\) from \(0 \leq t \leq x\) around the x-axis to the cylinder that it "sits" in.

   \[ \begin{align} \pi \int_0^x g(t)^2 dt &= \pi \int_0^x (at^3)^2 dt\\ &= \pi a^2 \int_0^x t^6 dt\\ &= \pi a^2 \left[\frac{t^7}{7}\right]_0^x\\ &= \frac{\pi}{7}a^2x^7\\\\ C_2(x) &= \pi x \int_0^x g(x)^2 dt\\ &= \pi x \cdot (ax^3)^2\\ &= \pi a^2 x^7\\\\ \frac{V(x)}{C_2(x)} &= \frac{\frac{\pi}{7}a^2x^7}{\pi a^2 x^7}\\ &= \frac{1}{7} \end{align} \]

4. Let \(h(t) = a \sqrt{t}\). Do the same thing as the previous question.

   \[ \begin{align} \pi \int_0^x h(t)^2 dt &= \pi \int_0^x (a\sqrt{t})^2 dt\\ &= \pi a^2 \int_0^x t dt\\ &= \pi a^2 \left[\frac{t^2}{2}\right]_0^x\\ &= \frac{\pi}{2}a^2x^2\\\\ C_3(x) &= \pi x \int_0^x h(x)^2 dt\\ &= \pi x \cdot (a\sqrt{x})^2\\ &= \pi a^2 x^2\\\\ \frac{V(x)}{C_3(x)} &= \frac{\frac{\pi}{2}a^2x^2}{\pi a^2 x^2}\\ &= \frac{1}{2} \end{align} \]

5. Show that for any power function of the form \(f(t) = at^k\), where \(a\) and \(k\) are positive real numbers, that the ratio of the volume and the cylinder that it "sits" in is always a constant.

   \[ \begin{align} \pi \int_0^x f(t)^2 dt &= \pi \int_0^x (at^k)^2 dt\\ &= \pi a^2 \int_0^x t^{2k} dt\\ &= \pi a^2 \left[\frac{t^{2k+1}}{2k+1}\right]_0^x\\ &= \frac{\pi}{2k+1}a^2x^{2k+1}\\\\ C_4(x) &= \pi x \int_0^x f(x)^2 dt\\ &= \pi x \cdot (ax^k)^2\\ &= \pi a^2 x^{2k+1}\\\\ \frac{V(x)}{C_4(x)} &= \frac{\frac{\pi}{2k+1}a^2x^{2k+1}}{\pi a^2 x^{2k+1}}\\ &= \frac{1}{2k+1} \end{align} \]

Let \(V(x)\) be the volumes that you found before and \(C(x)\) be the cylinders that you used before. You've shown \(\frac{V(x)}{C(x)}\) should be constant for power functions. Now we're going to prove a more interesting result.

1. Let \(y = f\) be a positive, increasing, twice-differentiable function. Let \(\frac{V(x)}{C(x)} = Q\), which is a constant Q. Differentiate both sides to show that \(V = \frac{cv'}{c'}\)

   \[ \begin{align} \frac{d}{dx}\left[\frac{V(x)}{C(x)}\right] &= \frac{d}{dx}Q\\ \frac{C(x)V'(x)-V(x)C'(x)}{C(x)^2} &= 0\\ C(x)V'(x)-V(x)C'(x) &= 0\\ C(x)V'(x) &= V(x)C'(x)\\\\ V(x) &= \frac{C(x)V'(x)}{C'(x)} \quad \blacksquare \end{align} \]

2. Use \(C(x)=\pi x\cdot f^2(x)\) and \(V(x) = \int_0^x \pi f^2(t) dt\). What is \(V'\)?

   \[ \begin{align} \frac{d}{dx}V(x) &= \frac{d}{dx}\int_0^x \pi f^2(t) dt\\ &= \pi f^2(x) \end{align} \]

3. Let \(y=f(x)\). If \(V = \frac{cv'}{c'}\), show by implicit differentiation that

   \[ \begin{align} y^2= \frac{(y+2xy')(3xy^2y'+y^3)-(xy^3)(3y'+2xy'')}{(y+2xy')^2}\\\\ \end{align} \]
   \[ \begin{align} V'(x) &= (\frac{cv'}{c'})'\\ &= \frac{c'[c'v' + cv''] - cv'c''}{(c')^2}\\\\ C'(x) &= (\pi y^2) + 2\pi xyy'\\ &= \pi y(y + 2xy')\\\\ C''(x) &= 2\pi yy' + 2\pi yy' + 2 \pi x (y')^2 + 2\pi x yy''\\ &= 2\pi (2 yy' + x(y')^2 + xy'')\\\\ V'(x) &= \pi y^2\\ V''(x) &= 2\pi yy'\\\\ \end{align} \]
   \[ \begin{align} \small y^2 = \frac{(y+2xy')[(y+2xy')(\pi y) + (\pi xy^2)(2\pi yy') - (\pi x y^2)(\pi y^2)(2\pi) (2 yy' + x(y')^2 + xy''))(y)]}{(y+2xy')^2} \end{align} \]
   \[ \begin{align} y^{2}&=\frac{y^{4}+2xy^{3}y^{\prime}+6x^{2}y^{2}(y)^{2}-2x^{2}y^{3}y^{\prime\prime}}{(y+2xy^{\prime})^{2}}\\\\ &= (y+2xy^{\prime})(3xy^{2}y^{\prime}+y^{3})-(xy^{3})(3y+2xy^{\prime\prime}) \\ &=3xy^{3}y^{\prime}+y^{4}+6x^{2}y^{2}(y^{\prime})^{2}+2xy^{3}y^{\prime} -3xy3y'-2x^2y^3y''\\ &=y^{4}+2xy^{3}y^{\prime}+6x^{2}y^{2}(y^{\prime})^{2}-2x^{2}y^{5}y^{\prime\prime}\\ &\text{ Expansion is equivalent to the original equation} \quad \blacksquare \end{align} \]

4. Now show that the above simplifies to \(xyy'' - xy'^2=-yy'\)

   \[ \begin{align} y^2 &= \frac{3xy^3y' + y^4 + 6x^2y^2(y')^2 + 2xy'y^3 - 2x^2y^3y''}{y^2 + 4xyy' + 4x^2(y')^2}\\ 1 &= \frac{y^2 + 6x^2(y')^2 + 2xy'y - 2x^2y''}{y^2 + 4xyy' + 4x^2(y')^2}\\\\ y^2 + 4xyy' + 4x^2(y')^2 &= y^2 + 6x^2(y')^2 + 2xy'y - 2x^2y''\\ 2yy' + 2x^2y'' &= 2x^2y''\\ xyy' -x(y')^2 &= -yy' \quad \blacksquare \end{align} \]

5. Divide both sides by \(xyy'\) and then integrate. (Hint: If you're confused, use a u-sub)

   \[ \begin{align} \frac{xyy''}{xyy'} - \frac{x(y')^2}{xyy'} &= -\frac{yy'}{xyy'}\\ \frac{y''}{y'} - \frac{y'}{y} &= -\frac{1}{x}\\ \int \frac{y''}{y'} - \frac{y'}{y} dx &= -\int \frac{1}{x} dx\\ \ln|y'| - \ln|y| + C &= -\ln|x|\\ \end{align} \]

6. Solve the result for \(\frac{y'}{y}\) and then integrate again. You should have shown something interesting about power functions.

   \[ \begin{align} \ln|y'| - \ln|y| &= -\ln|x| + C\\ \ln\left|\frac{y'}{y}\right| &= -\ln|x| + C\\ \frac{y'}{y} &= e^{-\ln|x| + C}\\ \frac{y'}{y} &= e^C \cdot e^{-\ln|x|}\\ \frac{y'}{y} &= e^C \cdot \frac{1}{x}\\ \int \frac{y'}{y} &= \int e^C \cdot \frac{1}{x} dx\\ \ln|y| &= e^C \cdot \ln|x| + D\\ \ln|y| &= \ln|x|^{e^C} + D\\ y &= x^D \cdot x^{e^C}\\ \end{align} \]

   The result gives us a power function.