10.3 Applications to area and volume

Area and Volume Calculation

Calculating Area Using Double Integrals

The simplest application of a double integral is finding the area of a region $R$ in the $xy$-plane. The idea: if you integrate the constant function 1 over a region, you get its area.

$\text{Area} = \iint_{R} dA$

The area element $dA$ depends on your coordinate system:

• Rectangular coordinates: $dA = dy\,dx$ (or $dx\,dy$)
• Polar coordinates: $dA = r\,dr\,d\theta$

Choosing the right order of integration matters. If the region's top and bottom boundaries are functions of $x$ (like $y = g_1(x)$ to $y = g_2(x)$), integrate with respect to $y$ first. If the left and right boundaries are functions of $y$, integrate with respect to $x$ first. Always sketch the region before deciding.

Example: Find the area enclosed between $y = x^2$ and $y = x$ for $0 \le x \le 1$.

1. The curves intersect at $x = 0$ and $x = 1$.
2. On this interval, $x \ge x^2$, so $y$ ranges from $x^2$ to $x$.
3. Set up the integral:

$\text{Area} = \int_0^1 \int_{x^2}^{x} dy\,dx = \int_0^1 (x - x^2)\,dx = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$

Switch to polar coordinates when the region has circular symmetry. For instance, the area of a disk of radius $a$:

$\text{Area} = \int_0^{2\pi}\int_0^{a} r\,dr\,d\theta = \pi a^2$

Volume Calculation Using Double Integrals

To find the volume of a solid that sits above a region $R$ in the $xy$-plane and below a surface $z = f(x,y) \ge 0$, you integrate the height function over $R$:

$\text{Volume} = \iint_{R} f(x,y)\,dA$

Think of it this way: $f(x,y)\,dA$ is the volume of a thin rectangular column at $(x,y)$ with height $f(x,y)$ and base area $dA$. The double integral sums all these columns.

Setting up the integral step by step:

1. Sketch the region $R$ in the $xy$-plane (the "shadow" of the solid).
2. Identify the boundary curves and decide on the order of integration.
3. Write the inner limits as functions of the outer variable.
4. The integrand is $f(x,y)$.
5. Evaluate inside-out.

Example: Find the volume under $z = 4 - x^2 - y^2$ above the $xy$-plane.

The surface meets the $xy$-plane where $4 - x^2 - y^2 = 0$, i.e., $x^2 + y^2 = 4$. The region $R$ is a disk of radius 2. Polar coordinates are the natural choice:

$V = \int_0^{2\pi}\int_0^{2}(4 - r^2)\,r\,dr\,d\theta = \int_0^{2\pi}\left[2r^2 - \frac{r^4}{4}\right]_0^2 d\theta = \int_0^{2\pi} 4\,d\theta = 8\pi$

If the solid is sandwiched between two surfaces $z = f(x,y)$ (top) and $z = g(x,y)$ (bottom), the volume becomes:

$V = \iint_R \bigl[f(x,y) - g(x,y)\bigr]\,dA$

Surface Area

For a surface $z = f(x,y)$ defined over a region $R$, the surface area accounts for how much the surface tilts away from horizontal. The formula is:

$\text{Surface Area} = \iint_{R} \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2}\,dA$

The expression under the square root equals 1 when the surface is flat (both partials are zero), which makes sense: surface area would just equal the area of $R$. The steeper the surface, the larger the partials, and the greater the surface area relative to $R$.

Steps to compute surface area:

1. Compute $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.
2. Plug into the formula under the square root.
3. Set up the double integral over $R$ with appropriate limits.
4. Evaluate (switching to polar coordinates often simplifies things for circular regions).

Physical Properties

Density Functions and Total Mass

A density function $\rho(x,y)$ gives the mass per unit area at each point in a region. If density is constant, $\rho(x,y) = \rho_0$. If it varies (say, a plate that's denser near its center), $\rho$ is a function of position.

The total mass of a lamina (thin flat plate) occupying region $R$ with density $\rho(x,y)$ is:

$M = \iint_{R} \rho(x,y)\,dA$

This is the foundation for every other physical property below. When $\rho(x,y) = 1$ (uniform, unit density), the mass integral reduces to the area integral.

Center of Mass

The center of mass $(\bar{x}, \bar{y})$ is the balance point of the lamina. You compute it by taking "weighted averages" of position, weighted by density:

$\bar{x} = \frac{1}{M}\iint_{R} x\,\rho(x,y)\,dA \qquad \bar{y} = \frac{1}{M}\iint_{R} y\,\rho(x,y)\,dA$

where $M = \iint_{R} \rho(x,y)\,dA$ is the total mass.

The numerator integrals are called the first moments: $M_y = \iint_R x\,\rho\,dA$ (moment about the $y$-axis) and $M_x = \iint_R y\,\rho\,dA$ (moment about the $x$-axis). A common source of confusion: $M_y$ uses $x$ in the integrand, not $y$, because it measures how far mass is distributed from the $y$-axis.

For a region with uniform density, the center of mass is called the centroid, and $\rho$ cancels out of the formulas.

Moments of Inertia

The moment of inertia quantifies how much a lamina resists rotational acceleration about a given axis. Mass farther from the axis contributes more (note the squared distance terms):

• About the $x$-axis: $I_x = \iint_{R} y^2\,\rho(x,y)\,dA$
• About the $y$-axis: $I_y = \iint_{R} x^2\,\rho(x,y)\,dA$
• About the origin (polar moment): $I_0 = I_x + I_y = \iint_{R} (x^2 + y^2)\,\rho(x,y)\,dA$

The polar moment $I_0$ is just the sum of the other two. In polar coordinates, $x^2 + y^2 = r^2$, which often simplifies the computation significantly for circular or annular regions.

Advanced Geometry

Curved Surfaces and Parametric Surface Area

When a surface can't be written as $z = f(x,y)$, you can parametrize it with a vector-valued function:

$\vec{r}(u,v) = \langle x(u,v),\, y(u,v),\, z(u,v) \rangle$

where $(u,v)$ ranges over a parameter domain $D$ in the $uv$-plane.

The surface area of the parametrized surface is:

$\text{Area} = \iint_{D} \left\|\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}\right\| du\,dv$

The cross product $\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}$ gives a normal vector to the surface, and its magnitude tells you how much a small rectangle in the $uv$-plane gets "stretched" when mapped onto the surface.

Steps to compute parametric surface area:

1. Find $\frac{\partial \vec{r}}{\partial u}$ and $\frac{\partial \vec{r}}{\partial v}$.
2. Compute their cross product.
3. Take the magnitude of the cross product.
4. Integrate that magnitude over the parameter domain $D$.

This generalizes the earlier surface area formula. In fact, if $z = f(x,y)$, you can set $\vec{r}(x,y) = \langle x, y, f(x,y)\rangle$, and the cross product magnitude reduces to $\sqrt{1 + f_x^2 + f_y^2}$, recovering the formula from the previous section.

More general surface integrals of a function $f(x,y,z)$ over a surface $S$ take the form:

$\iint_{S} f(x,y,z)\,dS = \iint_{D} f\bigl(\vec{r}(u,v)\bigr)\left\|\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}\right\| du\,dv$

These become essential in later units when you work with flux integrals and Stokes' theorem.