Integrals of Vector Functions
Component-wise integration, displacement, and work along a path - Kaplan §4.5
A particle spirals through a force field. How much work does the field do on it?
Vector-Valued Functions of One Variable
A vector function $\mathbf{r}(t) = x(t)\,\mathbf{i} + y(t)\,\mathbf{j} + z(t)\,\mathbf{k}$ traces a space curve as $t$ varies. Its derivative $\mathbf{r}'(t)$ is the velocity vector, tangent to the curve.
The velocity vector $\mathbf{r}'(t) = -\sin t\,\mathbf{i} + \cos t\,\mathbf{j} + \frac{1}{2\pi}\,\mathbf{k}$ always points in the direction of motion. Its magnitude $|\mathbf{r}'(t)|$ is the speed.
Component-Wise Integration
To integrate a vector function, we integrate each component separately:
For $\mathbf{r}'(t) = (-\sin t,\;\cos t,\;\tfrac{1}{2\pi})$, integrating from $0$ to $T$:
Notice the fundamental theorem for vector functions: $\int_0^T \mathbf{r}'(t)\,dt = \mathbf{r}(T) - \mathbf{r}(0)$, the net displacement.
Work as a Line Integral
If a force $\mathbf{F}$ acts on a particle moving along a path $\mathbf{r}(t)$, the work done is:
The integrand $\mathbf{F} \cdot \mathbf{r}'(t)$ is the component of force along the direction of motion. Positive when the field pushes with the motion (red), negative when it pushes against (blue).
Force field: $\mathbf{F}(x,y,z) = (-y,\; x,\; 1)$. Along the helix $\mathbf{r}(t) = (\cos t, \sin t, t/(2\pi))$, the dot product $\mathbf{F}\cdot\mathbf{r}' = \sin^2 t + \cos^2 t + \frac{1}{2\pi} = 1 + \frac{1}{2\pi}$, so $W = (1 + \frac{1}{2\pi}) \cdot 2\pi = 2\pi + 1 \approx 7.28$.
The work integral formally belongs to Kaplan §5.4, not §4.5 - we include this card here because it falls out naturally from component-wise integration. For the dedicated treatment with parametric reduction, conservative-field tests, and Kaplan worked problems, see §5.4 Line Integrals as Integrals of Vectors.
Playground - Custom Vector Integral
Enter a vector field and a parametric path, then watch the integral accumulate.
Practice Problems - §4.5
From Kaplan, problems after §4.5
Find the volume below $z = x^2 e^{-x-y}$ for $0 \le x \le 1$, $0 \le y \le 2$.
Step 1: Separate the integral
$$V = \int_0^1\int_0^2 x^2 e^{-x-y}\,dy\,dx = \left(\int_0^1 x^2 e^{-x}\,dx\right)\left(\int_0^2 e^{-y}\,dy\right)$$
Step 2: The $y$-integral
$$\int_0^2 e^{-y}\,dy = -e^{-y}\big|_0^2 = 1 - e^{-2}$$
Step 3: The $x$-integral by parts (twice)
Apply integration by parts with $u = x^2$, $dv = e^{-x}\,dx$:
$$\int_0^1 x^2 e^{-x}\,dx = \left[-x^2 e^{-x}\right]_0^1 + 2\int_0^1 x e^{-x}\,dx = -e^{-1} + 2\int_0^1 x e^{-x}\,dx$$
Apply parts again with $u = x$, $dv = e^{-x}\,dx$:
$$\int_0^1 x e^{-x}\,dx = \left[-xe^{-x}\right]_0^1 + \int_0^1 e^{-x}\,dx = -e^{-1} + (1-e^{-1}) = 1-2e^{-1}$$
So $\displaystyle\int_0^1 x^2 e^{-x}\,dx = -e^{-1} + 2(1 - 2e^{-1}) = 2 - 5e^{-1}$.
Step 4: Combine
$$\boxed{V = (2 - 5e^{-1})(1 - e^{-2})}$$
For $R$: $y^2 + x(x-1) \le 0$, write $\displaystyle\iint_R f\,dA$ as iterated integrals of both forms.
Step 1: Understand the region
Rewrite the inequality: $y^2 \le x - x^2 = x(1-x)$. This requires $x(1-x) \ge 0$, so $0 \le x \le 1$, and $|y| \le \sqrt{x(1-x)}$.
Geometrically, $x^2 - x + y^2 \le 0$ can be written $(x-\tfrac{1}{2})^2 + y^2 \le \tfrac{1}{4}$, which is a disk of radius $\tfrac{1}{2}$ centered at $(\tfrac{1}{2}, 0)$.
Step 2: Form 1 ($dy\,dx$)
For each $x \in [0,1]$, $y$ ranges from $-\sqrt{x(1-x)}$ to $\sqrt{x(1-x)}$:
$$\int_0^1\int_{-\sqrt{x(1-x)}}^{\sqrt{x(1-x)}} f\,dy\,dx$$
Step 3: Form 2 ($dx\,dy$)
The maximum of $\sqrt{x(1-x)}$ is $\tfrac{1}{2}$ at $x = \tfrac{1}{2}$, so $y$ ranges over $[-\tfrac{1}{2}, \tfrac{1}{2}]$.
For fixed $y$, solve $x^2 - x + y^2 \le 0$ for $x$: by the quadratic formula, $x \in \left[\frac{1-\sqrt{1-4y^2}}{2},\;\frac{1+\sqrt{1-4y^2}}{2}\right]$.
$$\int_{-1/2}^{1/2}\int_{(1-\sqrt{1-4y^2})/2}^{(1+\sqrt{1-4y^2})/2} f\,dx\,dy$$
Set up (do not evaluate) the mass of a sphere of radius $a$ whose density is proportional to the distance from one diametral plane.
Step 1: Choose coordinates and density
Take the diametral plane to be $z = 0$. Then the density is $\delta = k|z|$ for some constant $k$. The sphere is $x^2+y^2+z^2 \le a^2$.
Step 2: Set up in spherical coordinates
In spherical coordinates: $x = \rho\sin\phi\cos\theta$, $y = \rho\sin\phi\sin\theta$, $z = \rho\cos\phi$, so $|z| = \rho|\cos\phi|$ and $dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$.
Set up the moment of inertia about the $x$-axis of the solid $0 \le z \le 1 - x^2 - y^2$, $0 \le x \le 1$, $0 \le y \le 1-x$, with density $\delta = kxy$.
Step 1: Recall the formula
The moment of inertia about the $x$-axis is $I_x = \iiint_R (y^2+z^2)\,\delta\,dV$, where the factor $(y^2+z^2)$ is the squared distance from the $x$-axis.
Step 2: Set up the integral
Evaluate $\displaystyle\int_0^1 \mathbf{F}(t)\,dt$, where $\mathbf{F}(t) = t^2\,\mathbf{i} - e^t\,\mathbf{j} + \frac{1}{t+1}\,\mathbf{k}$.
Step 1: Integrate component by component
The integral of a vector function is just the vector of integrals of each component. We compute each one separately:
$$\int_0^1 \mathbf{F}(t)\,dt = \left(\int_0^1 t^2\,dt\right)\mathbf{i} - \left(\int_0^1 e^t\,dt\right)\mathbf{j} + \left(\int_0^1 \frac{1}{t+1}\,dt\right)\mathbf{k}$$
Step 2: Evaluate each integral
$\mathbf{i}$-component: $\displaystyle\int_0^1 t^2\,dt = \frac{t^3}{3}\bigg|_0^1 = \frac{1}{3}$
$\mathbf{j}$-component: $\displaystyle\int_0^1 e^t\,dt = e^t\big|_0^1 = e - 1$
$\mathbf{k}$-component: $\displaystyle\int_0^1 \frac{1}{t+1}\,dt = \ln(t+1)\big|_0^1 = \ln 2$
Step 3: Assemble the result
$$\boxed{\int_0^1 \mathbf{F}(t)\,dt = \frac{1}{3}\,\mathbf{i} - (e-1)\,\mathbf{j} + \ln 2\,\mathbf{k}}$$
Evaluate $\displaystyle\iint_R \mathbf{F}(x,y)\,dA$, where $R$ is the triangular region with vertices $(0,0)$, $(1,0)$, $(0,1)$ and $\mathbf{F}(x,y) = x^2 y\,\mathbf{i} + xy^2\,\mathbf{j}$.
Step 1: Set up the region and split into components
The triangle has vertices $(0,0)$, $(1,0)$, $(0,1)$. The hypotenuse is $x + y = 1$, so for each $x\in[0,1]$, $y$ ranges from $0$ to $1-x$.
We integrate each component of $\mathbf{F}$ separately over $R$:
$$\iint_R \mathbf{F}\,dA = \left(\iint_R x^2 y\,dA\right)\mathbf{i} + \left(\iint_R xy^2\,dA\right)\mathbf{j}$$
Step 2: Evaluate the $\mathbf{i}$-component
$$\int_0^1\int_0^{1-x} x^2 y\,dy\,dx = \int_0^1 x^2\cdot\frac{y^2}{2}\bigg|_0^{1-x}dx = \int_0^1 \frac{x^2(1-x)^2}{2}\,dx$$
Expanding: $x^2(1-x)^2 = x^2 - 2x^3 + x^4$, so
$$\frac{1}{2}\int_0^1 (x^2 - 2x^3 + x^4)\,dx = \frac{1}{2}\left(\frac{1}{3} - \frac{1}{2} + \frac{1}{5}\right) = \frac{1}{2}\cdot\frac{1}{30} = \frac{1}{60}$$
Step 3: Evaluate the $\mathbf{j}$-component
$$\int_0^1\int_0^{1-x} xy^2\,dy\,dx = \int_0^1 x\cdot\frac{y^3}{3}\bigg|_0^{1-x}dx = \int_0^1 \frac{x(1-x)^3}{3}\,dx$$
Expanding: $x(1-x)^3 = x - 3x^2 + 3x^3 - x^4$, so
$$\frac{1}{3}\int_0^1 (x - 3x^2 + 3x^3 - x^4)\,dx = \frac{1}{3}\left(\frac{1}{2} - 1 + \frac{3}{4} - \frac{1}{5}\right) = \frac{1}{3}\cdot\frac{1}{20} = \frac{1}{60}$$
Step 4: Assemble the result
By symmetry (swapping $x \leftrightarrow y$ swaps the two components and leaves $R$ unchanged), both components are equal. The answer is:
$$\boxed{\iint_R \mathbf{F}\,dA = \frac{1}{60}\,\mathbf{i} + \frac{1}{60}\,\mathbf{j}}$$