Volume of a Solid of Revolution

Method 1: Geometric Formula

Because $y=x$ forms a triangle, revolving it creates a right circular cone with a radius of 2 and a height of 2. Using the geometric formula:

$$ \text{Volume} = \frac{1}{3} \pi (\text{radius})^2 (\text{height}) = \frac{1}{3} \pi (2)^2 (2) = \frac{8\pi}{3} $$

Method 2: Definite Integral (Disk Method)

Let $f(x) = x$. Using Formula 1, the volume is given by the definite integral from 0 to 2:

$$ \text{Volume} = \int_{0}^{2} \pi x^2 \, dx = \left[ \pi \frac{x^3}{3} \right]_0^2 = \frac{8\pi}{3} $$

The graph of $y = \sqrt{r^2 - x^2}$ is bounded from $x = -r$ to $x = r$. The volume is given by the disk method (Formula 1):

$$ \text{Volume} = \int_{-r}^{r} \pi \left(\sqrt{r^2 - x^2}\right)^2 \, dx = \int_{-r}^{r} \pi (r^2 - x^2) \, dx $$

Find the antiderivative and evaluate at the limits:

$$ = \pi \left[ r^2 x - \frac{x^3}{3} \right]_{-r}^r $$ $$ = \pi \left[ \left(r^3 - \frac{r^3}{3}\right) - \left(-r^3 + \frac{r^3}{3}\right) \right] = \frac{4}{3} \pi r^3 $$

This derives the well-known formula for the volume of a sphere!

Since the solid is generated by revolving around the y-axis, we must rewrite our equations as functions of $y$ ($x = z(y)$). The limits of integration along the y-axis are from $y = 0$ to $y = 1$ (where the lines intersect).

The right curve is $y = -x + 2 \implies x = 2 - y$.

The left curve is $y = x \implies x = y$.

Using the washer method (Formula 4):

$$ \text{Volume} = \int_{0}^{1} \pi [ (2-y)^2 - y^2 ] \, dy $$

Expand the binomial and simplify the integrand:

$$ = \int_{0}^{1} \pi [ 4 - 4y + y^2 - y^2 ] \, dy = \int_{0}^{1} \pi [ 4 - 4y ] \, dy $$

Integrate and evaluate:

$$ = \pi \left[ 4y - 2y^2 \right]_0^1 = \pi (4 - 2) = 2\pi $$

We use the washer method along the x-axis (Formula 2). Let $f(x) = 1 + x^2$ (the outer radius) and $h(x) = \sqrt{x}$ (the inner radius).

$$ \text{Volume} = \int_{0}^{2} \pi ( f(x)^2 - h(x)^2 ) \, dx $$ $$ = \int_{0}^{2} \pi \left( (1+x^2)^2 - (\sqrt{x})^2 \right) \, dx $$

Expand the squares:

$$ = \pi \int_{0}^{2} ( 1 + 2x^2 + x^4 - x ) \, dx $$

Integrate using the power rule:

$$ = \pi \left[ x + \frac{2x^3}{3} + \frac{x^5}{5} - \frac{x^2}{2} \right]_0^2 $$

Evaluate at $x = 2$:

$$ = \pi \left( 2 + \frac{16}{3} + \frac{32}{5} - 2 \right) = \frac{176\pi}{15} $$

The equation of the circle is given by $x^2 + (y - R)^2 = r^2$.

Solve the equation for $y$ to obtain the upper and lower bounds of the circle:

• Upper semicircle: $f(x) = R + \sqrt{r^2 - x^2}$
• Lower semicircle: $h(x) = R - \sqrt{r^2 - x^2}$

Because of the symmetry with respect to the y-axis, we can integrate from $x = 0$ to $x = r$, then double the answer to find the total volume. Using the washer method (Formula 2):

$$ \frac{1}{2} \text{Volume} = \int_{0}^{r} \pi [ f(x)^2 - h(x)^2 ] \, dx $$ $$ = \pi \int_{0}^{r} \left[ (R + \sqrt{r^2 - x^2})^2 - (R - \sqrt{r^2 - x^2})^2 \right] \, dx $$

When you expand and subtract the squares, the $R^2$ and $(r^2 - x^2)$ terms cancel out, leaving:

$$ = 4 R \pi \int_{0}^{r} \sqrt{r^2 - x^2} \, dx $$

The integral of $\sqrt{r^2 - x^2}$ yields the area of a quarter circle, which we can solve using standard trigonometric substitution rules:

$$ = 4 R \pi \left( \frac{1}{2} \right) \left[ x \sqrt{r^2-x^2} + r^2\arcsin\left(\frac{x}{r}\right) \right]_0^r $$ $$ = 2 R \pi \left( r^2 \arcsin(1) - 0 \right) = 2 R \pi \left( r^2 \frac{\pi}{2} \right) = \pi^2 R r^2 $$

Since this was only half the torus, the total volume is twice the above result:

$$ \text{Volume} = 2 \pi^2 R r^2 $$

(Note: A minor typographical error in the original calculus derivation for the arcsin term has been corrected here for mathematical accuracy).

Use the washer method around the x-axis:

$$ \text{Volume} = \pi \int_{0}^{1} \left[ (x^2 + 2)^2 - (x)^2 \right] \, dx $$ $$ = \pi \int_{0}^{1} (x^4 + 4x^2 + 4 - x^2) \, dx = \pi \int_{0}^{1} (x^4 + 3x^2 + 4) \, dx $$

Integrate and evaluate at the bounds:

$$ = \pi \left[ \frac{x^5}{5} + x^3 + 4x \right]_0^1 = \pi \left( \frac{1}{5} + 1 + 4 \right) = \frac{26\pi}{5} $$

First, find the points of intersection by setting $x^3 = x^2$. The curves intersect at $x=0$ and $x=1$, which correspond to $y=0$ and $y=1$.

Since we are rotating around the y-axis, we need to express $x$ as functions of $y$. In the interval $y \in [0,1]$, $y = x^3 \implies x = y^{1/3}$ is the outer radius, and $y = x^2 \implies x = y^{1/2}$ is the inner radius.

Use the washer method around the y-axis:

$$ \text{Volume} = \pi \int_{0}^{1} \left[ (y^{1/3})^2 - (y^{1/2})^2 \right] \, dy = \pi \int_{0}^{1} (y^{2/3} - y) \, dy $$

Integrate and evaluate:

$$ = \pi \left[ \frac{3}{5}y^{5/3} - \frac{y^2}{2} \right]_0^1 = \pi \left( \frac{3}{5} - \frac{1}{2} \right) = \frac{\pi}{10} $$