Double integrals: regions and changing order

Find the vertices by intersecting pairs of boundary lines. At a fixed height $y$, solve $y=x+1$ for $x$ and decide which side of that line belongs to the triangle.

1. Describe the geometry. The two nonvertical lines meet at $(0,1)$. At $x=3$, they meet the vertical boundary at heights $1$ and $4$. Thus the horizontal extent is $0\le x\le3$, while the vertical extent is $1\le y\le4$.

2. Use vertical slices. For each fixed $x$ in that horizontal extent, $y$ starts at $1$ and ends at $x+1$. Therefore

The inner integral measures the slice's height, $(x+1)-1=x$. Hence

3. Use horizontal slices. Solve the sloping boundary as $x=y-1$. For a fixed $y$ between $1$ and $4$, the triangle begins at that line and ends at $x=3$. This gives

The width is $3-(y-1)=4-y$, so

4. Check a slice and the area. At $y=3$, the correct horizontal slice runs from $x=2$ to $x=3$. The triangle's base and perpendicular height are both $3$, giving $A=3(3)/2=9/2$.

Proposed setup: $\int_1^4\int_0^3 1\,dx\,dy$.

Diagnosis: It covers a rectangle. For example, $(0,3)$ is included even though it lies above $y=x+1$. This setup gives area $9$, double the triangle's area.

Repair: Replace the lower inner bound $0$ by $y-1$. After reversing order, a boundary that was the top of a vertical slice becomes the left end of a horizontal slice.

Use $x^2=2x$ to find the intersection coordinates. For reversed bounds, rewrite both boundary equations in terms of $x$; then use a test height to order the endpoints.

1. Find the bounded region. Factoring $x^2-2x=x(x-2)$ gives $x=0$ and $x=2$. Their heights are $0$ and $4$. On $0\le x\le2$, the difference $2x-x^2=x(2-x)$ is nonnegative, so the line is the upper boundary.

2. Set up vertical slices. The lower and upper heights are $x^2$ and $2x$:

Integrate the height:

3. Reverse the order. The entire bounded region lies in $x\ge0$, so the relevant inverse of the parabola is $x=\sqrt y$. The line becomes $x=y/2$. For $0\le y\le4$, the line is on the left; for example, at $y=1$, $1/2<1$. The endpoints meet again at $y=4$.

Therefore

4. Check the endpoints. At heights $0$ and $4$, the horizontal slice has zero width. At height $1$, its width is $1/2>0$. These checks help detect an inverted bound.

Proposed reversed bounds: $0\le y\le4$, $-\sqrt y\le x\le\sqrt y$.

Diagnosis: These bounds include both branches of the parabola and ignore the line. The point $(-1,1)$ would be included, although the bounded region between the two stated curves has no negative $x$-coordinates.

Repair: Keep the positive branch as the right boundary and use $x=y/2$ as the left boundary. The square-root sign comes from the region's location, not from a rule that every parabola contributes both roots.

Rewrite $x=y/2$ and $x=3-y$ as equations for $y$. The region must lie below both resulting lines. Where do those two possible upper boundaries meet?

1. Read the original bounds. The height ranges from $0$ to $2$. At height $y$, a slice starts at $x=y/2$ and ends at $x=3-y$. At $y=0$, these are $0$ and $3$. At $y=2$, they both equal $1$. The region is the triangle shown.

2. Find the change in the roof. The side equations are $y=2x$ and $y=3-x$. They meet where $2x=3-x$, giving $x=1$, $y=2$. From $x=0$ to $x=1$, a vertical slice reaches $y=2x$ first. From $x=1$ to $x=3$, it reaches $y=3-x$ first. Both slices start at $y=0$.

3. Write the two pieces. Let $I_L$ and $I_R$ denote the integrals over the left and right pieces. Then

The shared line $x=1$ has zero area; including it as the endpoint of both pieces does not add extra area to the double integral.

4. Evaluate each piece. An antiderivative of $1+y$ with respect to $y$ is $y+y^2/2$. Therefore

For the right piece, let $u=3-x$. This is just a one-variable substitution after the inner integral; as $x$ runs from $1$ to $3$, $u$ decreases from $2$ to $0$, and $du=-dx$. Thus

5. Check in the original order. The horizontal width is $3-3y/2$. Hence the original integral reduces to

Proposed setup: $\int_0^3\int_0^{3-x}(1+y)\,dy\,dx$.

Diagnosis: It uses the right roof everywhere and includes points above the left roof. For example, $(1/2,2)$ passes these proposed bounds but violates $y\le2x$. The incorrect value is $9$.

Repair: Split at $x=1$. Use $2x$ for the upper height on the left and $3-x$ on the right. If you use a minimum or a piecewise upper bound instead, you have encoded the same split rather than removed it.

The inequality $x^2\le y$ gives $-\sqrt y\le x\le\sqrt y$. You must also keep $x\ge-1$. At which height does $-\sqrt y$ pass to the left of $-1$?

1. Read the vertical description. The lower boundary is the parabola and the upper boundary is the horizontal line. The area is

For the last calculation, the antiderivative $4x-x^3/3$ has values $16/3$ at $2$ and $-11/3$ at $-1$.

2. Find the full height range. The lowest point is $(0,0)$; the top lies on $y=4$. Thus $0\le y\le4$. The height is not restricted to $1\le y\le4$ just because the left vertical edge starts at height $1$.

3. Compare the two left restrictions. For $0\le y\le1$, the value $-\sqrt y$ is at least $-1$, so the parabola controls the left endpoint. For $1\le y\le4$, the vertical line $x=-1$ is farther right and controls it. The right endpoint is $x=\sqrt y$ in both pieces; because $y\le4$, it already satisfies $x\le2$.

4. Write and evaluate the pieces. Let $A_{\mathrm{bottom}}$ and $A_{\mathrm{top}}$ denote the areas below and above height $1$, respectively.

5. Check representative slices. At $y=1/4$, the endpoints are $-1/2$ and $1/2$. At $y=9/4$, they are $-1$ and $3/2$. Both descriptions agree with the parabola and the clipping line.

Proposed reversed bounds: $0\le y\le4$, $-\sqrt y\le x\le\sqrt y$.

Diagnosis: They forget the clipping line. For example, $(-3/2,3)$ satisfies these proposed bounds and lies above the parabola, but violates $x\ge-1$. The proposed area is $32/3$, larger than $9$.

Repair: Keep the parabola's negative branch only up to $y=1$, then replace the left endpoint by $-1$. You must preserve every original restriction when changing order.

Because $x$ and $y$ are nonnegative here, $\sqrt x\le y$ is equivalent to $x\le y^2$. If you integrate with respect to $x$ first, the exponential is constant along each slice.

1. Describe horizontal slices. The region starts at height $0$, ends at height $1$ and has left endpoint $x=0$. Its right endpoint at height $y$ is $x=y^2$. The original restriction $x\le1$ is automatic because $0\le y\le1$.

2. Write the reversed integral. No split is needed: the same left and right boundaries work throughout the full height range.

3. Integrate in the easy variable first. Holding $y$ fixed, the exponential has no $x$-dependence. Its inner integral is its value multiplied by the slice's width:

Thus

4. Substitute. Let $u=y^3$, so $du=3y^2\,dy$. The new bounds remain $0$ and $1$. Therefore

There is no need to find an elementary antiderivative of the original inner integrand with respect to $y$. The new inner integral supplies precisely the $y^2$ factor needed for the substitution.

5. Check scale. The region's area is $\int_0^1 y^2\,dy=1/3$. On it, $1\le e^{y^3}\le e$, so

The answer is approximately $0.57276$, within those bounds.

Proposed reversed upper bound: $x=\sqrt y$.

Diagnosis: This repeats a square root rather than solving the original boundary for $x$. At $y=1/2$, the correct right endpoint is $x=1/4$, whereas the proposed endpoint is about $0.707$. The extra points do not satisfy $y\ge\sqrt x$.

Repair: Rewrite $y=\sqrt x$ as $x=y^2$, using the nonnegative coordinates to justify squaring. Verify by substitution at one height before evaluating anything.

For vertical slices with $0\le x\le1$, the inner circle creates a lower boundary. For $1\le x\le2$, the slice starts on the horizontal axis. In polar coordinates, $x^2+y^2=r^2$; check that the bounds refer to $r$, not $r^2$.

1. Describe Cartesian slices. Define the upper-circle height

and the inner-circle height

For $0\le x\le1$, the slice runs from $a(x)$ to $b(x)$. For $1\le x\le2$, it runs from $0$ to $b(x)$. Let $J_1$ and $J_2$ be the contributions from these two horizontal ranges:

These are valid bounds, but they introduce two pieces and square roots.

2. Describe polar slices. The radial coordinate $r$ is the distance to the origin; the angle $\theta$ is measured counterclockwise from the positive $x$-axis in radians. The constant $\pi$ has its usual circle value. Here

At every permitted angle, the entire slice runs between the same two radii, with no split.

3. Transform both the integrand and the area element. The integrand becomes $f=r^2$. A narrow polar cell has radial thickness $dr$ and tangential width approximately $r\,d\theta$; the exact change-of-variables formula uses

The extra $r$ accounts for the increasing area swept out by the same angle farther from the origin. Thus

4. Evaluate. The radial integral is

It is independent of $\theta$, so

5. Check size. The area is one quarter of the difference of the disk areas:

The average value of $f$ is $J/A=5/2$, between its minimum $1$ and maximum $4$ on the region. This is a plausibility check, not a replacement for the setup.

Proposed polar setup: $\int_0^{\pi/2}\int_1^2 r^2\,dr\,d\theta$.

Diagnosis: The radii and angles are correct, but the polar area factor is missing. This gives $7\pi/6$, which computes a different weighted integral.

Repair: Transform $dA$ along with the function. Multiply $r^2$ by $r$ to obtain $r^3$. Also note that the inequality $1\le r^2\le4$ gives $1\le r\le2$, because $r\ge0$.