Multivariable-Calculus · Unit 14 · Video 2 · Interactive Practice

Double Integrals: Volume Under a Surface by Slicing

IKey Formulas

Formula	Name	Meaning
$\displaystyle\iint_R f(x,y)\,dA = \lim_{\Delta A \to 0} \sum_i f(x_i, y_i)\,\Delta A_i$	Definition	Volume under the graph (for $f \ge 0$ ) as a limit of thin columns
$\displaystyle\text{Volume} = \int_{x_{\min}}^{x_{\max}} S(x)\,dx$	Adding slices	$S(x)$ is the area of the slice at $x$
$\displaystyle S(x) = \int_{y_{\min}(x)}^{y_{\max}(x)} f(x,y)\,dy$	Area of one slice	$x$ held fixed, $y$ is the variable
$\displaystyle\iint_R f\,dA = \int_{x_{\min}}^{x_{\max}} \int_{y_{\min}(x)}^{y_{\max}(x)} f(x,y)\,dy\,dx$	Iterated integral	Inner integral first, then outer

Key Insight: the outer bounds are numbers (the first and last slice); the inner bounds depend on the outer variable and record the shape of $R$ .

IIVisualization 1 — Thin Columns Approach the Volume

Under $z = 2 - x^2 - y^2$ over the unit square, the column sums converge to the volume $\tfrac{4}{3}$ .

Grid n × n 2 Sample point

IIIVisualization 2 — Slice the Solid, Then Add the Slices

Slices $x = x_0$ of the dome $z = 1 - \tfrac{x^2}{4} - y^2$ over the oval $R$ have areas $S(x_0)$ that add up to its volume.

Slice x₀ 0.80

IVVisualization 3 — Outer Bounds Are Numbers, Inner Bounds Move

In either order, the inner bounds follow the region's boundary as the outer variable changes.

Region Order Slice at 1.20

VQuiz Questions

Problem 1 · Iterated Integral over a Rectangle

Given: $R$ is the rectangle $0 \le x \le 1$ , $0 \le y \le 2$ — find $\displaystyle\int_0^1 \int_0^2 (x + 2y)\,dy\,dx$ .

✅ Correct! The inner integral gives

S(x) = 2x + 4

, and

\int_0^1 (2x + 4)\,dx = 1 + 4 = 5

❌ Bounds swapped.

4

\int_0^2 \int_0^1 (x + 2y)\,dy\,dx

: the inner bounds belong to

y

(from

0

2

), the outer bounds to

x

(from

0

1

❌ Check the $x$ term. With

x

held fixed,

\int_0^2 x\,dy = x \cdot (2 - 0) = 2x

, not

x

❌ Wrong range for $y$ .

\tfrac{3}{2}

integrates

y

from

0

1

; the rectangle has

0 \le y \le 2

Show solution

Inner integral ( $x$ fixed, $y$ from $0$ to $2$ ):

S(x) = \int_0^2 (x + 2y)\,dy = \Big[\,xy + y^2\,\Big]_{y=0}^{y=2} = 2x + 4

Outer integral:

\int_0^1 (2x + 4)\,dx = \Big[\,x^2 + 4x\,\Big]_0^1 = 1 + 4 = 5

The volume under the plane $z = x + 2y$ above $R$ is $5$ .

Problem 2 · Setting Up the Bounds

Given: $R$ is the region between the curves $y = x^2$ and $y = x$ — which iterated integral equals $\iint_R f(x,y)\,dA$ in the order $dy\,dx$ ?

✅ Correct! The curves meet at

x = 0

and

x = 1

(the outer bounds), and for

0 \le x \le 1

the slice runs from

y = x^2

up to

y = x

❌ Outer bounds must be numbers. After the outer integration over

x

nothing may still depend on

x

; bounds that contain

x

belong on the inner

dy

integral.

❌ Lower and upper are swapped. For

0 \le x \le 1

we have

x^2 \le x

, so

y

runs from

x^2

x

; this version gives the negative of the answer.

❌ That is the whole unit square. For a fixed

x

y

does not run from

0

1

inside

R

; the inner bounds must follow the two curves.

Show solution

Outer bounds (first and last slice): the curves meet where $x^2 = x$ , i.e. $x = 0$ and $x = 1$ .

Inner bounds: a vertical segment at a fixed $x$ in $(0, 1)$ enters $R$ on the lower curve $y = x^2$ and leaves it on the upper curve $y = x$ (for instance at $x = \tfrac12$ : $\tfrac14 \le y \le \tfrac12$ ).

\iint_R f\,dA = \int_0^1 \int_{x^2}^{x} f(x,y)\,dy\,dx

Problem 3 · Slice Area, Then Volume

Given: $R$ is the region between $y = x^2$ and $y = x$ , and $f(x,y) = x$ — find the slice area $S(x)$ and the volume $\iint_R x\,dA$ .

What is S(x)?

What is the volume?

✅ Correct!

S(x) = x(x - x^2) = x^2 - x^3

, and

\int_0^1 (x^2 - x^3)\,dx = \tfrac13 - \tfrac14 = \tfrac{1}{12}

❌ That is only the length of the segment.

x - x^2

\int_{x^2}^{x} 1\,dy

; the integrand here is

f = x

, a constant in

y

, so the slice area is

x \cdot (x - x^2)

❌ Check the lower bound.

x^2

\int_0^{x} x\,dy

; inside

R

the slice starts at

y = x^2

, not at

y = 0

❌ Lower and upper are swapped. On

0 \le x \le 1

x^2 \le x

, so the integral runs from

x^2

up to

x

and

S(x)

is positive.

❌ That is the area of $R$ .

\tfrac16 = \int_0^1 (x - x^2)\,dx

integrates

1

, not

f = x

, over

R

❌ The lower curve was dropped.

\tfrac13 = \int_0^1 x^2\,dx

is the volume over the triangle

0 \le y \le x

, not over the region above

y = x^2

❌ A volume under a positive integrand cannot be negative.

-\tfrac{1}{12}

comes from integrating

y

downward, from

x

x^2

❌ Not quite. Hold

x

fixed and integrate

x

with respect to

y

from

x^2

x

❌ Not quite. Integrate

S(x) = x^2 - x^3

from

0

1

Show solution

Bounds: outer $0 \le x \le 1$ (where $x^2 = x$ ); inner $x^2 \le y \le x$ .

Inner integral ( $x$ is a constant here):

S(x) = \int_{x^2}^{x} x\,dy = x\,\big[\,y\,\big]_{y=x^2}^{y=x} = x(x - x^2) = x^2 - x^3

Outer integral:

\int_0^1 (x^2 - x^3)\,dx = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}

Problem 4 · Bounds and Volume over a Triangle

Given: $R$ is the triangle with vertices $(0,0)$ , $(2,0)$ and $(0,2)$ , bounded by the axes and the line $y = 2 - x$ , and $f(x,y) = y$ — set up $\iint_R y\,dA$ in the order $dy\,dx$ and find it.

Which iterated integral?

What is the volume?

✅ Correct! The slices run from

x = 0

x = 2

, each from

y = 0

up to the line

y = 2 - x

; then

S(x) = \tfrac{(2-x)^2}{2}

and

\int_0^2 \tfrac{(2-x)^2}{2}\,dx = \tfrac{4}{3}

❌ That is the whole square $[0,2] \times [0,2]$ . At a fixed

x

the slice stops at the line

y = 2 - x

, not at

y = 2

; the inner upper bound must depend on

x

❌ That is the other half of the square.

2 - x \le y \le 2

is the triangle above the line, with vertices

(2,0)

(2,2)

(0,2)

;

R

lies below the line, from

y = 0

up to

y = 2 - x

❌ The inner bound uses the wrong variable. The bounds of a

dy

integral cannot contain

y

, the variable being integrated; solve the line for

y

, giving

y = 2 - x

❌ That is the area of $R$ .

2 = \int_0^2 (2 - x)\,dx

integrates

1

over

R

; the integrand here is

f = y

❌ That is the volume over the other half of the square.

\tfrac{8}{3} = \int_0^2 \int_{2-x}^{2} y\,dy\,dx

; over

R

itself the inner integral runs from

0

2 - x

❌ That is the volume over the whole square.

4 = \int_0^2 \int_0^2 y\,dy\,dx

; the slice at

x

ends at

y = 2 - x

❌ Not quite. The outer bounds are the first and last slice,

x = 0

and

x = 2

; at a fixed

x

y

runs from the

x

-axis up to the line.

❌ Not quite. Integrate

S(x) = \int_0^{2-x} y\,dy

from

0

2

Show solution

Outer bounds (first and last slice): $R$ spans $0 \le x \le 2$ .

Inner bounds: a vertical segment at a fixed $x$ enters $R$ on the $x$ -axis, $y = 0$ , and leaves it on the line $y = 2 - x$ (for instance at $x = \tfrac12$ : $0 \le y \le \tfrac32$ ).

\iint_R y\,dA = \int_0^2 \int_0^{2-x} y\,dy\,dx

Inner integral ( $x$ fixed):

S(x) = \int_0^{2-x} y\,dy = \Big[\,\tfrac{y^2}{2}\,\Big]_{y=0}^{y=2-x} = \frac{(2-x)^2}{2}

Outer integral:

\int_0^2 \frac{(2-x)^2}{2}\,dx = \Big[\,-\frac{(2-x)^3}{6}\,\Big]_0^2 = 0 + \frac{8}{6} = \frac{4}{3}

Check: over the whole square $\int_0^2 \int_0^2 y\,dy\,dx = 4$ , and the triangle above the line holds $\tfrac83$ ; indeed $\tfrac43 + \tfrac83 = 4$ .

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