Double integrals over general regions — Type I and II: Exercises

Throughout, remember the golden rule from the parent:

Level 1 — Recognition

Here we only set up the limits or read them off a picture. No hard algebra yet.

Recall Solution

The outer variable is $x$ (its limits $0$ and $2$ are pure numbers). The inner variable $y$ runs between two functions of $x$ : bottom curve $g_{1} (x) = x^{2}$ , top curve $g_{2} (x) = 2 x$ . Because $y$ is squeezed between curves of $x$ , this is a Type I region — we sweep vertical strips.

Look at the figure: at a fixed $x$ (say $x = 1$ ), the strip starts on the parabola $y = x^{2} = 1$ and ends on the line $y = 2 x = 2$ . Every vertical strip has its feet on $y = x^{2}$ and its head on $y = 2 x$ .

Figure — Double integrals over general regions — Type I and II

(Sanity check that $2 x \geq x^{2}$ on $[0, 2]$ : at $x = 1$ , $2 \geq 1$ ✓; they meet at $x = 0$ and $x = 2$ .)

Recall Solution

The three vertices give three edges. The slanted edge joins $(4, 0)$ and $(0, 2)$ ; its line is $\frac{x}{4} + \frac{y}{2} = 1$ , i.e. $y = 2 - \frac{x}{2}$ .

For vertical strips: $x$ ranges over the shadow of the triangle on the $x$ -axis, from $0$ to $4$ (numbers ✓). For fixed $x$ , the strip's foot is on the $x$ -axis $y = 0$ and its head is on the slant $y = 2 - \frac{x}{2}$ . $\int_{0}^{4} \int_{0}^{2 - x /2} x y d y d x .$ We stop here — L1 asks only for the setup.

Level 2 — Application

Now we set up and evaluate.

Recall Solution

The slanted edge joins $(0, 0)$ and $(1, 2)$ : line $y = 2 x$ . The right edge is $x = 1$ ; the bottom is $y = 0$ .

Draw the vertical strip (figure below): for fixed $x \in [0, 1]$ , $y$ climbs from the $x$ -axis $y = 0$ to the slant $y = 2 x$ .

$\int_{0}^{1} \int_{0}^{2 x} (x + y) d y d x .$ Inner (treat $x$ as constant): $\int_{0}^{2 x} (x + y) d y = [x y + \frac{y ^{2}}{2}]_{0}^{2 x} = x (2 x) + \frac{( 2 x ) ^{2}}{2} = 2 x^{2} + 2 x^{2} = 4 x^{2} .$ Outer: $\int_{0}^{1} 4 x^{2} d x = \frac{4}{3} .$ So the integral equals $\frac{4}{3}$ .

Recall Solution

Now sweep horizontal strips. The $y$ -shadow of the triangle is $0 \leq y \leq 2$ (numbers ✓). For a fixed height $y$ , the strip's left boundary is the slant $y = 2 x \Rightarrow x = \frac{y}{2}$ , and its right boundary is $x = 1$ . $\int_{0}^{2} \int_{y /2}^{1} (x + y) d x d y .$ Inner: $\int_{y /2}^{1} (x + y) d x = [\frac{x ^{2}}{2} + y x]_{y /2}^{1} = (\frac{1}{2} + y) - (\frac{y ^{2}}{8} + \frac{y ^{2}}{2}) = \frac{1}{2} + y - \frac{5 y ^{2}}{8} .$ Outer: $\int_{0}^{2} (\frac{1}{2} + y - \frac{5 y ^{2}}{8}) d y = [\frac{y}{2} + \frac{y ^{2}}{2} - \frac{5 y ^{3}}{24}]_{0}^{2} = 1 + 2 - \frac{40}{24} = 3 - \frac{5}{3} = \frac{4}{3} .$ ✅ Same value $\frac{4}{3}$ — the value is order-independent; only the algebra differs.

Recall Solution

The curves meet where $x = x \Rightarrow x = x^{2} \Rightarrow x = 0, 1$ . Test $x = \frac{1}{4}$ : $x = \frac{1}{2} > \frac{1}{4} = x$ , so $x$ is on top. $\int_{0}^{1} \int_{x}^{x} 2 y d y d x .$ Inner: $\int_{x}^{x} 2 y d y = [y^{2}]_{x}^{x} = x - x^{2} .$ Outer: $\int_{0}^{1} (x - x^{2}) d x = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} .$ Result $\frac{1}{6}$ .

Level 3 — Analysis

Here you must choose the smart order, or split the region.

Recall Solution

$\int sin (y^{3}) d y$ has no elementary antiderivative, so the given ( $d y$ -first) order is a dead end. We must switch order.

Read the region. Given: $0 \leq x \leq 1$ and $x \leq y \leq 1$ . Since $y \geq x$ means $x \leq y^{2}$ , and $x \geq 0$ , for a fixed $y$ the variable $x$ runs from $0$ to $y^{2}$ . The $y$ -shadow is $0 \leq y \leq 1$ .

Rewrite as Type II: $\int_{0}^{1} \int_{0}^{y^{2}} sin (y^{3}) d x d y .$ Inner ( $sin (y^{3})$ is constant in $x$ ): $\int_{0}^{y^{2}} sin (y^{3}) d x = y^{2} sin (y^{3}) .$ Outer: let $u = y^{3}, d u = 3 y^{2} d y$ : $\int_{0}^{1} y^{2} sin (y^{3}) d y = \frac{1}{3} \int_{0}^{1} sin u d u = \frac{1}{3} (1 - cos 1) .$ Result $\frac{1 - c o s 1}{3} \approx 0.1533.$

Why it worked: switching turned the impossible inner integral into a trivial one, and the leftover $y^{2}$ is exactly the derivative factor a $u = y^{3}$ substitution needs.

Recall Solution

Meet where $y^{2} = y + 2 \Rightarrow y^{2} - y - 2 = 0 \Rightarrow (y - 2) (y + 1) = 0 \Rightarrow y = - 1, 2$ . As a Type II region (horizontal strips), for each $y \in [- 1, 2]$ : which curve is left, which is right? Test $y = 0$ : $x = y^{2} = 0$ (parabola), $x = y + 2 = 2$ (line). Parabola is left, line is right. $Area = \int_{- 1}^{2} \int_{y^{2}}^{y + 2} 1 d x d y = \int_{- 1}^{2} ((y + 2) - y^{2}) d y .$ $= [\frac{y ^{2}}{2} + 2 y - \frac{y ^{3}}{3}]_{- 1}^{2} = (2 + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3}) = \frac{10}{3} - (- \frac{7}{6}) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} .$ Area $= \frac{9}{2}$ .

Why Type II? As Type I you'd have to split at $x = 1$ (below $x = 1$ both walls are the parabola; above, the top wall becomes the line) — two integrals. Type II sweeps the whole thing in one shot. This is the same area idea as Area between two curves.

Level 4 — Synthesis

Combine region-building with a physical quantity.

Recall Solution

Curves meet at $x = 0, 1$ ; on $(0, 1)$ , $x^{2} < x$ so $y = x$ is top. Type I: $m = \int_{0}^{1} \int_{x^{2}}^{x} x d y d x = \int_{0}^{1} x (x - x^{2}) d x = \int_{0}^{1} (x^{2} - x^{3}) d x = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} .$ Mass $= \frac{1}{12}$ . (This is the setup used in Centre of mass and moments.)

Recall Solution

$M_{y} = \int_{0}^{1} \int_{x^{2}}^{x} x^{2} d y d x = \int_{0}^{1} x^{2} (x - x^{2}) d x = \int_{0}^{1} (x^{3} - x^{4}) d x = \frac{1}{4} - \frac{1}{5} = \frac{1}{20} .$ With $m = \frac{1}{12}$ from L4·1: $\overset{x}{ˉ} = \frac{M _{y}}{m} = \frac{1/20}{1/12} = \frac{12}{20} = \frac{3}{5} .$ So $M_{y} = \frac{1}{20}$ and $\overset{x}{ˉ} = \frac{3}{5}$ .

Level 5 — Mastery

Everything at once: choose the order, handle a genuinely two-piece region, verify.

Recall Solution

$\int \frac{e ^{x}}{x} d x$ is non-elementary — the $d x$ -first order fails. Switch.

Region: $0 \leq y \leq 2$ and $\frac{y}{2} \leq x \leq 1$ . From $x \geq \frac{y}{2}$ we get $y \leq 2 x$ ; with $x \leq 1$ the $x$ -shadow is $0 \leq x \leq 1$ , and for fixed $x$ , $y$ runs $0$ to $2 x$ . $\int_{0}^{1} \int_{0}^{2 x} \frac{e ^{x}}{x} d y d x .$ Inner ( $\frac{e ^{x}}{x}$ constant in $y$ ): $\int_{0}^{2 x} \frac{e ^{x}}{x} d y = \frac{e ^{x}}{x} \cdot 2 x = 2 e^{x} .$ Outer: $\int_{0}^{1} 2 e^{x} d x = 2 (e - 1) .$ Result $2 (e - 1) \approx 3.4366.$ Why it worked: the strip length $2 x$ cancelled the offending $\frac{1}{x}$ , leaving a textbook exponential.

Recall Solution

Inner: $\int_{0}^{4 - x^{2}} x d y = x 4 - x^{2} .$ Outer: substitute $u = 4 - x^{2}, d u = - 2 x d x$ . When $x = 1, u = 3$ ; when $x = 2, u = 0$ . $I = \int_{1}^{2} x 4 - x^{2} d x = - \frac{1}{2} \int_{3}^{0} u d u = \frac{1}{2} \int_{0}^{3} u d u = \frac{1}{2} \cdot \frac{2}{3} u^{3/2}_{0}^{3} = \frac{1}{3} \cdot 3^{3/2} = 3 .$ So $I = 3 \approx 1.732.$ (For the full quarter-disc $x$ from $0$ to $2$ the same method gives $\int_{0}^{2} x 4 - x^{2} d x = \frac{1}{3} \cdot 8 = \frac{8}{3}$ . In polar this integral is often cleaner — see Polar coordinates double integrals.)

Recall Solution

The region is the triangle $x \geq 0, y \geq 0, x + y \leq 1$ , symmetric in $x \leftrightarrow y$ . First order: inner $\int_{0}^{1 - x} (x + y) d y = [x y + \frac{y ^{2}}{2}]_{0}^{1 - x} = x (1 - x) + \frac{( 1 - x ) ^{2}}{2} = \frac{1 - x ^{2}}{2} .$ Then $\int_{0}^{1} \frac{1 - x ^{2}}{2} d x = \frac{1}{2} (1 - \frac{1}{3}) = \frac{1}{3} .$ Second order is identical by the $x \leftrightarrow y$ symmetry of both the region and integrand, giving $\frac{1}{3}$ again. Common value $= \frac{1}{3}$ . This is Fubini's Theorem in action.

Recall One-line self-test (fold to check)

Which order kills a non-elementary inner integrand — and why? ::: Switch so the messy factor (e.g. $e^{y^{2}}, sin (y^{3}), e^{x} / x$ ) becomes a constant during the inner integration, and the strip length supplies the derivative factor the outer integral needs.