Lecture: Sec 15.1 Double Integrals over Rectangles

Topic 1: From Area to Volume 🚀

In Calculus I, you became an expert at using the definite integral to find the area under a curve. The core idea was to slice the area into thin vertical rectangles and add them up, taking a limit as their width approached zero.

$$ \text{Area} = \int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x $$

Now, we want to level up. Instead of a curve $y=f(x)$ over an interval $[a,b]$, we have a surface $z=f(x,y)$ over a rectangular domain in the xy-plane, $R = [a,b] \times [c,d]$. Our new question is: What is the volume of the solid that lies above the rectangle $R$ and below the surface $z=f(x,y)$?

Just as we used rectangles to approximate area, we will use rectangular boxes (or prisms) to approximate volume.

Topic 2: The Double Integral as a Limit

The Rigorous Definition (The "What")

To find this volume, we follow the same "slice and sum" strategy, but now in two dimensions.

  1. Partition the Domain: We divide the rectangle $R$ into a grid of smaller subrectangles, each with area $\Delta A = \Delta x \Delta y$.
  2. Pick a Sample Point: In each subrectangle $R_{ij}$, we choose a sample point $(x_{ij}^*, y_{ij}^*)$.
  3. Form the Prism: The value of the function at the sample point, $f(x_{ij}^*, y_{ij}^*)$, gives us the height of a rectangular prism. The volume of this single prism is its height times its base area: $f(x_{ij}^*, y_{ij}^*) \Delta A$.
  4. Sum the Volumes: We add up the volumes of all these small prisms to get an approximation of the total volume. This is called a double Riemann sum: $$ V \approx \sum_{i=1}^m \sum_{j=1}^n f(x_{ij}^*, y_{ij}^*) \Delta A $$
  5. Take the Limit: The exact volume is found by taking the limit as the number of subrectangles goes to infinity (making their size shrink to zero). This limit is the double integral. $$ V = \iint_R f(x,y) \, dA = \lim_{m,n \to \infty} \sum_{i=1}^m \sum_{j=1}^n f(x_{ij}^*, y_{ij}^*) \Delta A $$

Signed vs. Geometric Volume

Important: Just like "net signed area" in Calc I, if $f(x,y)$ is ever negative on $R$, then $\iint_R f(x,y)\,dA$ represents a net signed volume, where volume below the xy-plane is counted as negative. The actual geometric volume is $\iint_R |f(x,y)| \, dA$.

Topic 3: Iterated Integrals and Fubini's Theorem

The limit definition of the double integral is great for theory, but terrible for computation. Luckily, there's a much easier way, thanks to Guido Fubini. We can compute the volume by performing two single integrals, one after the other. This is called an iterated integral.

Fubini's Theorem

Hypotheses: If $f(x,y)$ is continuous on the rectangle $R = [a,b] \times [c,d]$, then the iterated integrals exist and are equal to the double integral:

$$ \iint_R f(x,y) \, dA = \int_c^d \left[ \int_a^b f(x,y) \, dx \right] dy = \int_a^b \left[ \int_c^d f(x,y) \, dy \right] dx. $$

Key takeaway: Fubini's Theorem converts a double integral (a 2D problem) into a sequence of two single integrals (two 1D problems), which we already know how to solve!

Visualizing Fubini's Theorem: The Slicing Method

Think of the inner integral as finding the area of a 2D cross-section of the solid. The outer integral then "sweeps" or "stacks" these cross-sectional areas together to find the total volume. Interact with the graph below to see this in action.

  • To slice parallel to the yz-plane, select the folder `x=k` and move the slider `a` to the right. This shows the area of one slice. Then move the slider `n` to the right to "stack" these slices and form the volume.
  • To slice parallel to the xz-plane, select the folder `y=l` and move slider `b` to the right, followed by slider `m`.

Common Pitfalls to Avoid

  • Inner integral: treat the other variable as constant. Example: $\displaystyle \int (x-3y^2)\,dx=\tfrac12 x^2-3y^2x$ (OK). $\displaystyle \int (x-3y^2)\,dx=\tfrac12 x^2-y^3$ (illegal: differentiated $y$ inside a $dx$ integral).
  • Bounds must match the variable. The bounds of the inner integral must match the inner differential (e.g., $dx$ goes with x-bounds).

Example 1: A Basic Polynomial

Evaluate $\iint_R (x - 3y^2) \, dA$ where $R = \{(x,y) \mid 0 \le x \le 2, 1 \le y \le 2\}$.

Solution (Order 1: dx dy):

We'll integrate with respect to $x$ first (the inner integral), treating $y$ as a constant.

$$ \int_1^2 \left[ \int_0^2 (x - 3y^2) \, dx \right] dy $$

Inner integral: $\int_0^2 (x - 3y^2) \, dx = \left[ \frac{1}{2}x^2 - 3y^2x \right]_{x=0}^{x=2} = (2 - 6y^2) - (0) = 2 - 6y^2$.

Outer integral: $\int_1^2 (2 - 6y^2) \, dy = \left[ 2y - 2y^3 \right]_1^2 = (4 - 16) - (2 - 2) = -12$.

Example 2: Changing the Order of Integration

Evaluate $\int_0^2 \int_1^2 (x - 3y^2) \, dy \, dx$ to confirm Fubini's Theorem.

Solution:

Inner integral: $\int_1^2 (x - 3y^2) \, dy = \left[ xy - y^3 \right]_{y=1}^{y=2} = (2x - 8) - (x - 1) = x - 7$.

Outer integral: $\int_0^2 (x - 7) \, dx = \left[ \frac{1}{2}x^2 - 7x \right]_0^2 = (2 - 14) - 0 = -12$.

The results match! The negative value of the integral means that more of the solid's volume is below the xy-plane than above it.

Check Your Understanding 🤔

Evaluate the double integral $\iint_R y \sin(xy) \, dA$ over the rectangle $R = [1,2] \times [0,\pi]$.

Solution:

It's easier to integrate with respect to $x$ first, since the $y$ outside will be treated as a constant.

$$ \int_0^\pi \left[ \int_1^2 y \sin(xy) \, dx \right] dy $$

Inner: $\int_1^2 y \sin(xy) \, dx = y \left[ -\frac{1}{y}\cos(xy) \right]_{x=1}^{x=2} = -[\cos(xy)]_{x=1}^{x=2} = -(\cos(2y) - \cos(y)) = \cos(y) - \cos(2y)$.

Outer: $\int_0^\pi (\cos(y) - \cos(2y)) \, dy = \left[ \sin(y) - \frac{1}{2}\sin(2y) \right]_0^\pi = (0 - 0) - (0 - 0) = 0$.

Topic 4: A Special Case: Separable Functions

If the integrand $f(x,y)$ can be factored into a function of $x$ times a function of $y$, and the domain is a rectangle, the double integral simplifies into a product of two single integrals. This is a powerful shortcut.

Theorem for Separable Functions

If $f(x,y) = g(x)h(y)$ is continuous on $R=[a,b]\times[c,d]$, then

$$ \iint_R g(x)h(y) \, dA = \left( \int_a^b g(x) \, dx \right) \left( \int_c^d h(y) \, dy \right) $$

Justification: Using Fubini's theorem, we can write $\int_c^d \int_a^b g(x)h(y) \, dx \, dy$. In the inner integral, $h(y)$ is a constant, so it can be pulled out: $\int_c^d h(y) \left( \int_a^b g(x) \, dx \right) dy$. The term $\int_a^b g(x)dx$ is just a number, so it can be pulled out of the outer integral, proving the result.

Example: Using the Separable Function Shortcut

Evaluate $\iint_R x^2y \, dA$ over the rectangle $R = [0,3] \times [1,2]$.

Solution:

The function $f(x,y) = x^2y$ is separable with $g(x)=x^2$ and $h(y)=y$. The domain is a rectangle. Therefore, we can write:

$$ \iint_R x^2y \, dA = \left( \int_0^3 x^2 \, dx \right) \left( \int_1^2 y \, dy \right) $$ $$ = \left[ \frac{1}{3}x^3 \right]_0^3 \cdot \left[ \frac{1}{2}y^2 \right]_1^2 = (9 - 0) \cdot \left(\frac{4}{2} - \frac{1}{2}\right) = 9 \cdot \frac{3}{2} = \frac{27}{2}. $$

Check Your Understanding 🤔

Set up the integral $\iint_R \sin(x)\cos(y) \, dA$ over $R=[0, \pi/2] \times [0, \pi/2]$ as a product of two single integrals, and then evaluate.

Solution:

$$ \left( \int_0^{\pi/2} \sin(x) \, dx \right) \left( \int_0^{\pi/2} \cos(y) \, dy \right) $$ $$ = \left( [-\cos(x)]_0^{\pi/2} \right) \cdot \left( [\sin(y)]_0^{\pi/2} \right) = (-(0-1)) \cdot (1-0) = 1 \cdot 1 = 1. $$

Lecture Conclusion

Today, we extended integration to two dimensions. We defined the double integral as a limit of Riemann sums, representing a signed volume. We then used Fubini's Theorem to compute these integrals using simpler iterated integrals and learned a powerful shortcut for separable functions over rectangular domains.

What You Should Have Learned

  • The definition of a double integral as the limit of a double Riemann sum.
  • How to use Fubini's Theorem to evaluate double integrals over rectangles.
  • How to apply the separable function shortcut as a powerful computational tool.

Final Practice Problems

Practice Problem #1

Set up, but do not solve, the two iterated integrals for finding the volume of the solid under the plane $z=2x+y$ and above the rectangle $R = [-1,1] \times [0,3]$.

Solution:

The volume is given by the double integral $\iint_R (2x+y) \, dA$.

Order 1 (dy dx): $\displaystyle \int_{-1}^1 \int_0^3 (2x+y) \, dy \, dx $

Order 2 (dx dy): $\displaystyle \int_0^3 \int_{-1}^1 (2x+y) \, dx \, dy $

Practice Problem #2

Find the volume of the solid lying under the surface $z = xe^{-y}$ and above the rectangle $R = [0,2] \times [0,1]$.

Solution:

The function $f(x,y)=xe^{-y}$ is separable and the domain is a rectangle. This is the fastest method.

$$ V = \left( \int_0^2 x \, dx \right) \left( \int_0^1 e^{-y} \, dy \right) = \left[ \frac{1}{2}x^2 \right]_0^2 \cdot \left[ -e^{-y} \right]_0^1 $$ $$ = (2) \cdot (-e^{-1} - (-e^0)) = 2(1 - e^{-1}) = 2 - \frac{2}{e}. $$

Practice Problem #3

Evaluate $\displaystyle\iint_R \frac{xy^2}{x^2+1} \, dA$ over the rectangle $R = \{ (x,y) \mid 0 \le x \le 1, -3 \le y \le 3 \}$.

Solution:

The integrand is separable: $f(x,y) = \frac{x}{x^2+1} \cdot y^2$. We can use the shortcut.

$$ \iint_R \frac{xy^2}{x^2+1} \, dA = \left( \int_0^1 \frac{x}{x^2+1} \, dx \right) \left( \int_{-3}^3 y^2 \, dy \right) $$

First integral (u-substitution): Let $u=x^2+1$, so $du=2x\,dx$.

$$ \int_0^1 \frac{x}{x^2+1} \, dx = \frac{1}{2} \int_1^2 \frac{1}{u} \, du = \frac{1}{2} [\ln|u|]_1^2 = \frac{1}{2}(\ln(2)-\ln(1)) = \frac{1}{2}\ln(2). $$

Second integral:

$$ \int_{-3}^3 y^2 \, dy = \left[ \frac{1}{3}y^3 \right]_{-3}^3 = \frac{1}{3}(27 - (-27)) = \frac{54}{3} = 18. $$

Final Answer: Multiply the results.

$$ \left( \frac{1}{2}\ln(2) \right) \cdot (18) = 9\ln(2). $$