Multivariable-Calculus · Unit 14 · Video 3 · Interactive Practice

Iterated Integrals Over a Square and a Quarter Disk

IKey Formulas

Formula	What it is	Value
$\displaystyle\int_0^1\!\!\int_0^1 (1 - x^2 - y^2)\,dy\,dx$	Unit square; slice area $\tfrac23 - x^2$	$\tfrac13$
$\displaystyle\int_0^1\!\!\int_0^{\sqrt{1-x^2}} (1 - x^2 - y^2)\,dy\,dx$	Quarter disk; slice area $\tfrac23(1 - x^2)^{3/2}$	$\tfrac{\pi}{8}$
$dA = dy\,dx$	Small rectangle $\Delta A = \Delta y\,\Delta x$ ; the $dy$ integral is done first	—
$\cos^4\theta = \tfrac38 + \tfrac12\cos 2\theta + \tfrac18\cos 4\theta$	Double-angle formula used twice	$\displaystyle\int_0^{\pi/2}\cos^4\theta\,d\theta = \tfrac{3\pi}{16}$

Key Insight: The inner bounds may depend on $x$ — they give the range of $y$ in one slice. The outer bounds are numbers — they give the first slice and the last one.

IIVisualization 1 — One Slice, Two Regions

For fixed $x$ , the inner integral is the signed area of one slice; only its top bound changes between regions.

Region Slice at x 0.50

💡 Over the square, the part of each slice above the plane has exactly the area of the quarter-disk slice, $\tfrac23(1-x^2)^{3/2}$ ; the square adds only the negative piece beyond the circle.

IIIVisualization 2 — Adding Up the Slices

The outer integral accumulates slice areas $S(x)$ from the first slice at $x = 0$ to the last at $x = 1$ .

Upper limit x 0.60

IVVisualization 3 — Why Only the $\tfrac38$ Survives

Substituting $x = \sin\theta$ keeps the area $\tfrac{3\pi}{16}$ ; in $\cos^4\theta$ only the constant $\tfrac38$ contributes.

Step 1

Step 1 — The outer integral

\int_0^1 \tfrac23(1 - x^2)^{3/2}\,dx

VQuiz Questions

Problem 1 · The Inner Integral

Given: $x$ is held constant. Find $\displaystyle\int_0^2 (3x^2 + y)\,dy$ .

✅ Correct! The constant

3x^2

integrates to

3x^2 y

, which is

6x^2

y = 2

, and

\tfrac12 y^2

gives

2

❌ Not quite.

3x^2

is a constant in

y

, so its antiderivative is

3x^2 y

, not

3x^2

; at

y = 2

it contributes

6x^2

❌ Not quite. The antiderivative of

y

\tfrac12 y^2

, which is

2

y = 2

, not

4

❌ Not quite. That integrates in

x

from

0

2

. The inner integral is in

y

, so the result depends on

x

and contains no

y

Show solution

With $x$ fixed, $3x^2$ is a constant and $y$ is the variable:

\int_0^2 (3x^2 + y)\,dy = \left[3x^2 y + \tfrac12 y^2\right]_0^2 = (6x^2 + 2) - 0 = 6x^2 + 2.

No $y$ is left, because $y$ was integrated out; $x$ remains, because the slice depends on $x$ .

Problem 2 · Bounds for a Curved Edge

Given: $R$ is the part of the disk $x^2 + y^2 \le 4$ in the first quadrant. Which iterated integral equals $\iint_R f\,dA$ , with $dy$ done first?

✅ Correct! In the slice at

x

y

runs from the

x

-axis up to the circle,

y = \sqrt{4 - x^2}

; the slices run from

x = 0

x = 2

❌ Not quite. Those are the bounds of the square

[0,2]\times[0,2]

, which also includes the corner outside the circle.

❌ Not quite. The outer bounds must be numbers — they name the first and last slice. After both integrals no

x

may be left over.

❌ Not quite. Solving

x^2 + y^2 = 4

for

y

gives

y = \sqrt{4 - x^2}

; the curve

y = 4 - x^2

is a parabola, not the circle.

Show solution

Inner bounds. Fix $x$ . The slice starts on the $x$ -axis, $y = 0$ , and ends on the circle: $x^2 + y^2 = 4 \Rightarrow y = \sqrt{4 - x^2}$ (positive root, first quadrant).

Outer bounds. The first slice is at $x = 0$ (full height $2$ ) and the last at $x = 2$ (height $0$ ).

\iint_R f\,dA = \int_0^2\!\!\int_0^{\sqrt{4 - x^2}} f(x,y)\,dy\,dx.

Problem 3 · The Paraboloid over a Triangle

Given: $R$ is the triangle $0 \le x \le 1$ , $0 \le y \le x$ . Find $\displaystyle\int_0^1\!\!\int_0^{x} (1 - x^2 - y^2)\,dy\,dx$ .

✅ Correct! The slice area is

x - \tfrac43 x^3

, and

\int_0^1 \left(x - \tfrac43 x^3\right)dx = \tfrac12 - \tfrac13 = \tfrac16

❌ Not quite. In the inner integral

x^2

is a constant, so it integrates to

x^2 y

, which is

x^3

y = x

— not

x^2

❌ Not quite. That is the unit-square answer: the top bound here is

y = x

, not

y = 1

❌ Not quite. The antiderivative of

y^2

\tfrac13 y^3

; dropping the

\tfrac13

gives slice area

x - 2x^3

, whose integral is

0

Show solution

Inner integral ( $x$ constant, top bound $y = x$ ):

\int_0^x (1 - x^2 - y^2)\,dy = \left[(1 - x^2)\,y - \tfrac13 y^3\right]_0^x = (1 - x^2)\,x - \tfrac13 x^3 = x - \tfrac43 x^3.

Outer integral:

\int_0^1 \left(x - \tfrac43 x^3\right)dx = \left[\tfrac12 x^2 - \tfrac13 x^4\right]_0^1 = \tfrac12 - \tfrac13 = \tfrac16.

Problem 4 · A Wider Paraboloid

Given: $\displaystyle\int_0^2\!\!\int_0^{\sqrt{4 - x^2}} (4 - x^2 - y^2)\,dy\,dx$ , whose inner integral is $\tfrac23(4 - x^2)^{3/2}$ . Find its value using $x = 2\sin\theta$ and $\int_0^{\pi/2}\cos^4\theta\,d\theta = \tfrac{3\pi}{16}$ .

✅ Correct!

\tfrac23 \cdot 8 \cdot 2 \cdot \tfrac{3\pi}{16} = 2\pi

❌ Not quite. With

x = 2\sin\theta

the differential is

dx = 2\cos\theta\,d\theta

; the factor

2

is missing.

❌ Not quite.

4 - x^2 = 4\cos^2\theta

, so

(4 - x^2)^{3/2} = 8\cos^3\theta

; the factor

8

is missing.

❌ Not quite. That is the radius-

1

answer. Both

(4 - x^2)^{3/2} = 8\cos^3\theta

and

dx = 2\cos\theta\,d\theta

carry factors of the radius.

Show solution

Substitute $x = 2\sin\theta$ : then $4 - x^2 = 4\cos^2\theta$ , so $(4 - x^2)^{3/2} = 8\cos^3\theta$ (with $\cos\theta \ge 0$ on $[0, \tfrac{\pi}{2}]$ ), and $dx = 2\cos\theta\,d\theta$ . The bounds $x: 0 \to 2$ become $\theta: 0 \to \tfrac{\pi}{2}$ .

\int_0^2 \tfrac23(4 - x^2)^{3/2}\,dx = \tfrac23\int_0^{\pi/2} 8\cos^3\theta \cdot 2\cos\theta\,d\theta = \tfrac{32}{3}\int_0^{\pi/2}\cos^4\theta\,d\theta = \tfrac{32}{3}\cdot\tfrac{3\pi}{16} = 2\pi.

Check: it is $16$ times $\tfrac{\pi}{8}$ , as expected when every length doubles and the height scales by $4$ : $2 \cdot 2 \cdot 4 = 16$ .

Solved: 0 / 4

Iterated Integrals Over a Square and a Quarter Disk

IKey Formulas

IIVisualization 1 — One Slice, Two Regions

IIIVisualization 2 — Adding Up the Slices

IVVisualization 3 — Why Only the 38\tfrac3883​ Survives

VQuiz Questions

IVVisualization 3 — Why Only the $\tfrac38$ Survives