Double integrals II

To err is human, to blame it on someone else is even more human, Jacob’s Law

Functions, Partials, and Gradients

• A function of two variables $f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ assigns to each ordered pair in its domain a unique real number. Examples include geometric formulas like Area = $\frac{1}{2}bh$, or algebraic surfaces like $z = f(x, y) = 2x + 3y$, $f(x, y) = x^2 + y^2$, and $f(x,y) = e^{x+y}$.

• Partial derivatives differentiate a multivariable function $f(x_1, x_2, \dots, x_n)$ with respect to one target variable while treating all other variables as constants. They measure how the function changes in a single specific direction.
  If $f(x, y) = 2x^2y^3$, then $\frac{\partial f}{\partial x} = f_x = 4xy^3$ and $\frac{\partial f}{\partial y} = f_y = 6x^2y^2$.
  If $f(x, y) = x^3y + y^2$, then $f_x = 3x^2y$ and $f_y = x^3+2y$.

• A point $(x_1, x_2, \dots, x_n)$ is considered a critical point if the gradient of $f$ at that point is the zero vector, or if the gradient does not exist.

Gradients, Level Surfaces, and Directional Derivatives

• The gradient $\nabla f$ is a vector containing all first-order partial derivatives of $f$. It geometrically points in the direction of the function’s steepest ascent. $\nabla f = \left\langle \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \dots, \frac{\partial f}{\partial x_n} \right\rangle$.

• In 2D space ($n = 2$), finding a critical point $(x_0, y_0)$ requires solving the system: $\frac{\partial f}{\partial x}(x_0, y_0) = 0$ and $\frac{\partial f}{\partial y}(x_0, y_0) = 0$.

• Theorem: The gradient vector is perpendicular (orthogonal) to the level curves or level surfaces of the function at any given point. This means if you walk along a contour line where the function’s value is constant, the direction of steepest uphill is exactly 90 degrees to your path.

The directional derivative $\frac{dw}{ds}\bigg\vert{}_{\vec{u}}$ represents the rate of change of a function $w$ as you move in the direction of the unit vector $\vec{u}$. It can be expressed or compactly written via the dot product of the gradient and the direction vector: $\frac{dw}{ds}\bigg\vert{}_{\vec{u}} = \nabla w \cdot \vec{u} = \vert{}\nabla w\vert{}\vert{}\vec{u}\vert{}\cos(\theta) = \vert{}\nabla w\vert{}\cos(\theta)$ where $\theta$ is the angle between the gradient $\nabla w$ and the unit vector $\vec{u}$.

A double integral allows us to integrate a function of two variables, say $f(x, y)$, over a region R in a two-dimensional space (typically the xy-plane). Instead of just finding or calculating a 2D area, with a double integral, we’re looking for the volume under the surface (or below the graph) $z = f(x, y)$ and above the region $R$.

• Fubini’s Theorem. Suppose $f(x, y)$ is a continuous function defined on a rectangular region $R = [a, b] \times [c, d]$ in the xy-plane. Fubini’s Theorem states that the double integral of f(x, y) over the region R can be computed as an iterated integral in either of two orders and yet get the exact same result: $\boxed{\int \int_R f(x, y)dA = \int_{a}^{b}\int_{c}^{d} f(x, y)dydx = \int_{c}^{d}\int_{a}^{b} f(x, y)dxdy}$

Exchanging order of integration

When evaluating a double integral over a region, the order of integration refers to the sequence in which we perform integration with respect to different variables, for example, $dxdy$ versus $dydx$.

For rectangular regions (where all limits are constants), exchanging the order is as simple as swapping the integration bounds: $\int_{0}^{1} (\int_{0}^{2} f(x, y)dx)dy = \int_{0}^{2} (\int_{0}^{1} f(x, y)dy)dx$ (Figure i), $\int_{0}^{2}(\int_{1}^{3} xy^2dy)dx = \int_{1}^{3} (\int_{0}^{2} xy^2dx)dy$.

However, for non-rectangular regions, swapping the order requires redefining the region entirely. You cannot simply swap the limits if they contain variables. This technique is incredibly powerful because it allows us to simplify calculations by choosing the most convenient order of integration based on the given problem—and in some cases, it is the only way to evaluate an integral that has no elementary antiderivative in its original order.

Solved exercises

• Compute the double integral: $\int_{0}^{1} \int_{\tan^{-1}(x)}^{\frac{\pi}{4}} \cos(y)dydx$

Compute the inner integral. $\int_{\tan^{-1}(x)}^{\frac{\pi}{4}} \cos(y)dy = \left[\sin(y)\right]_{\tan^{-1}(x)}^{\frac{\pi}{4}}$
$= \sin\left(\frac{\pi}{4}\right) - \sin(\tan^{-1}(x)) =[...] \frac{1}{\sqrt{2}} - \frac{x}{\sqrt{1+x^2}}$

[…] Evaluating $\sin(\tan^{-1}(x))$: Recall the right triangle definition of trigonometric functions. If we set $\theta = \tan^{-1}(x)$, it means $\tan(\theta) = \frac{x}{1} = \frac{\text{Opposite}}{\text{Adjacent}}$. Using the Pythagorean theorem, the hypotenuse is $\sqrt{1+x^2}$. Therefore, $\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{x}{\sqrt{1+x^2}}$(Figure C).

Compute the outer integral. $\int_{0}^{1} (\frac{1}{\sqrt{2}}-{\frac{x}{\sqrt{1+x^2}}})dx$
This last integral can be solved using a substitution method. Let $u = 1+x^2$, which means $du = 2xdx$ (or $xdx = \frac{du}{2}$).
$\int \frac{x}{\sqrt{1+x^2}}dx = \frac{1}{2} \int u^{-1/2}du = \frac{1}{2} (2u^{1/2}) = \sqrt{1+x^2}$
$\int_{0}^{1} (\frac{1}{\sqrt{2}}-{\frac{x}{\sqrt{1+x^2}}})dx = (\frac{1}{\sqrt{2}}x-\sqrt{1+x^2})\bigg|_{0}^{1} = \frac{1}{\sqrt{2}}-\sqrt{2}-(0-1) = \frac{1}{\sqrt{2}}-\frac{2}{\sqrt{2}}+1 = 1 -\frac{1}{\sqrt{2}}$

• Avoiding an Impossible Antiderivative. Compute the double integral: $\int_{0}^{1} \int_{y}^{1}\frac{1}{1+x^4}dxdy$

Change the order of integration to simplify the computation. Sometimes, changing the order of integration can simplify the computation significantly, or even make an impossible integral solvable. The original integral requires us to integrate $\frac{1}{1+x^4}$ with respect to $x$ first, which is highly complex. Instead, let's reverse the order. (Figure D).

Looking at the original limits, the region is bounded by $x = y$ on the left and $x = 1$ on the right, for $y$ ranging from $0$ to $1$. To change the order to $dydx$, we analyze the region from the $x$-axis perspective: $x$ ranges from $0$ to $1$, and for each fixed $x$, $y$ goes from the bottom boundary ($y = 0$) up to the diagonal line ($y = x$).

Thus, the double integral can be rewritten as: $\int_{0}^{1} \int_{y}^{1}\frac{1}{1+x^4}dxdy = \int_{0}^{1} \int_{0}^{x}\frac{1}{1+x^4}dydx.$

Compute the inner integral (with respect to y): Because the integrand contains only $x$, it acts as a constant relative to $y$, $\int_{0}^{x}\frac{1}{1+x^4}dy = \frac{1}{1+x^4}y\bigg|_{0}^{x} = \frac{x}{1+x^4}$.

Compute the outer integral (with respect to x): $\int_{0}^{1} \frac{x}{1+x^4}dx$
Use substitution: let $u = x^2$, so $du = 2xdx$ (and $xdx = \frac{du}{2}$).
$\frac{1}{2} \int_{0}^{1} \frac{1}{1+u^2} du = \frac{1}{2}\left[\tan^{-1}(u)\right]_{0}^{1}$

The standard integral $\int \frac{1}{1+u^2}du = \tan^{-1}(u)$ is a fundamental result in calculus worth memorizing.

• Compute the double integral: $\int_{0}^{1} \int_{\sqrt{x}}^{1}\cos(y^3)dydx$

Change the order of integration to simplify the computation. Sometimes, changing the order of integration can simplify the computation significantly. The function $\cos(y^3)$ has no elementary antiderivative, so we must change the order. The original region has $y$ ranging from the curve $y = \sqrt{x}$ up to $y = 1$, with $x$ ranging from $0$ to $1$. If we swap perspectives, $y$ ranges from $0$ to $1$ overall, and for a fixed $y$, $x$ ranges from $0$ to the curve $x = y^2$. (Figure 1).

1. Thus, the double integral can be rewritten as $\int_{0}^{1} \int_{0}^{y^2} \cos(y^3) dxdy$
2. Compute the inner integral (with respect to x): $\int_{0}^{y^2} \cos(y^3) dx = \left[x \cos(y^3)\right]_{0}^{y^2} = y^2\cos(y^3)$.
3. Compute the outer integral (with respect to y): $\int_{0}^{1} y^2\cos(y^3) dy$
   This integral can be evaluated using a substitution method, let $u = y^3$, so $du = 3y^2dy$ (or $y^2dy = \frac{du}{3}$). When $y = 0, u = 0$. When $y = 1, u = 1$.
   Thus, the integral becomes: $\frac{1}{3}\int_{0}^{1} \cos(u)du = \frac{1}{3} \left[\sin(u)\right]_{0}^{1} = \frac{1}{3}(\sin(1) - \sin(0)) = \frac{\sin(1)}{3}$

• Choosing the Easier Path via Fubini’s Theorem. Compute the double integral: $\int_R xye^{xy^2}dA$ over the rectangular region $R = [0, 1] \times [0, 1]$.

1. We are going to use Fubini’s Theorem. It states that given a multivariable continuous function f(x, y) defined on a rectangular region $R = [0, 1] \times [0, 1]$ in the xy-plane, the double integral of f(x, y) over the region R can be computed as an iterated integral in either of two orders: $\int \int_R f(x, y)dA = \int_{a}^{b}\int_{c}^{d} f(x, y)dydx = \int_{c}^{d}\int_{a}^{b} f(x, y)dxdy$
   In this case, we are asked to compute: $\int_R xye^{xy^2}dA = \int_{0}^{1}\int_{0}^{1} xye^{xy^2}dxdy = \int_{0}^{1}\int_{0}^{1} xye^{xy^2}dydx$. We can choose to compute the integral in either order. Here, we will compute the integral with respect to y (it is much easier because the $y$ term perfectly matches the derivative of the exponent $xy^2$).
2. Compute the inner integral (with respect to y): $\int_{0}^{1} xye^{xy^2}dy =$
   Use substitution: let $u = xy^2$, so $du = 2xydy$ (treating $x$ as a constant), meaning $\frac{1}{2}du = xydy$.
   $= \int_{y=0}^{y=1} \frac{1}{2} e^u du = \left[\frac{1}{2}e^{xy^2}\right]_{y=0}^{y=1} = \frac{1}{2}(e^{x(1)^2} - e^{x(0)^2}) = \frac{1}{2}(e^x - 1)$
3. Compute the outer integral (with respect to x): $\int_{0}^{1} \frac{1}{2}(e^x-1)dx = \frac{1}{2}(e^x-x)\bigg|_{0}^{1} = \frac{1}{2}[(e-1)-(1-0)] = \frac{1}{2}(e-2)$

• Compute the double integral: $\int_{0}^{8} \int_{\sqrt[3]{x}}^{2}\frac{1}{y^4+1}dydx$

Changed the order of integration to simplify the computation. Sometimes, changing the order of integration is the only way to proceed. The original integral requires us to integrate $\frac{1}{y^4+1}$ with respect to $y$ first, which has no elementary antiderivative. To bypass this, we sketch the region and swap the integration order to $dxdy$ (Figure 2).

1. Analyzing the original limits, $x$ ranges from $0$ to $8$, and $y$ ranges from the curve $y = \sqrt[3]{x}$ up to the horizontal line $y = 2$. To change the perspective, look at the $y$-axis: $y$ ranges from $0$ to $2$. For any fixed $y$, $x$ starts at the y-axis ($x = 0$) and goes to the curve $x = y^3$.
   Thus, the double integral can be rewritten by swapping the order of integration: $\int_{0}^{8} \int_{\sqrt[3]{x}}^{2}\frac{1}{y^4+1}dydx = \int_{0}^{2} \int_{0}^{y^3}\frac{1}{y^4+1}dxdy$
2. Compute the inner integral (with respect to x): $\int_{0}^{y^3}\frac{1}{y^4+1}dx = \frac{1}{y^4+1}x\bigg|_{0}^{y^3} = \frac{y^3}{y^4+1}$

Compute the outer integral (with respect to y) $\int_{0}^{2} \frac{y^3}{y^4+1} dy$.
This integral is a standard form, which can be solved using substitution. Let $u = y^4 + 1$, meaning $du = 4y^3dy$, or $y^3dy = \frac{du}{4}$.
Rewriting the integral in terms of u, we get: $\int_{0}^{2} \frac{y^3}{y^4+1} dy = \frac{1}{4}\int \frac{1}{u}du = \frac{1}{4}ln|u| = \frac{1}{4}ln|y^4+1|\bigg|_{0}^{2} = \frac{1}{4}(ln(17)-ln(1)) = \frac{ln(17)}{4}$.

• Compute the volume under the paraboloid $z = 1 - x^2 - y^2$ restricted to the first quadrant ($x \ge 0, y \ge 0$) where $x^2 + y^2 \le 1$.

1. Unlike a square boundary, this region (figure D) is a quarter disk. For a fixed $x_0$, $y$ varies from $0$ up to the boundary of the circle, $y = \sqrt{1-x_0^2}$.
   Therefore, $V = \int_{0}^{1}\left(\int_{0}^{\sqrt{1-x^2}} (1-x^2-y^2)dy\right)dx$

2. Compute the Inner Integral: (Integrate with respect to y, then evaluate the inner integral) $\int_{0}^{\sqrt{1-x^2}} (1-x^2-y^2)dy = \left[y - x^2y - \frac{y^3}{3}\right]_{0}^{\sqrt{1-x^2}}$
   Factor out $y$: $= \left[y(1 - x^2) - \frac{y^3}{3}\right]_{0}^{\sqrt{1-x^2}}$
   Substitute the upper limit: $= \sqrt{1-x^2}(1-x^2) - \frac{1}{3}(\sqrt{1-x^2})^3 = (1-x^2)^{\frac{3}{2}} - \frac{1}{3}(1-x^2)^{\frac{3}{2}} = \frac{2}{3}(1-x^2)^{\frac{3}{2}}$

3. Compute the Outer Integral (with respect to x): $\int_{0}^{1} \frac{2}{3}(1-x^2)^{\frac{3}{2}}dx$
   Use trigonometric substitution: Let $x = \sin(\theta)$. Then $dx = \cos(\theta)d\theta$. When $x=0, \theta=0$; when $x=1, \theta=\frac{\pi}{2}$.
   Besides, $(1-x^2)^{\frac{1}{2}} = (1-\sin^2(\theta))^{\frac{1}{2}} = \cos(\theta)$.
   $= \int_{0}^{\frac{\pi}{2}} \frac{2}{3}(\cos^2(\theta))^{\frac{3}{2}}\cos(\theta)d\theta = \frac{2}{3}\int_{0}^{\frac{\pi}{2}} \cos^4(\theta)d\theta$
   

   Power-Reducing Identities: $\cos^2(\theta) = \frac{1+\cos(2\theta)}{2}$, $\cos^4(\theta) = \left(\frac{1+\cos(2\theta)}{2}\right)^2 = \frac{1}{4}(1 + 2\cos(2\theta) + \cos^2(2\theta))$. Finally, we apply the identity again for $\cos^2(2\theta) = \frac{1+\cos(4\theta)}{2}$.

   Substituting the expanded identity: $\frac{2}{3} \cdot \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \left(1 + 2\cos(2\theta) + \frac{1}{2} + \frac{\cos(4\theta)}{2}\right)d\theta = \frac{1}{6} \int_{0}^{\frac{\pi}{2}} \left(\frac{3}{2} + 2\cos(2\theta) + \frac{\cos(4\theta)}{2}\right)d\theta$
   $= \frac{1}{6} \left[\frac{3}{2}\theta + \sin(2\theta) + \frac{1}{8}\sin(4\theta)\right]_{0}^{\frac{\pi}{2}} = \frac{1}{6}\left(\frac{3}{2}\cdot\frac{\pi}{2} + 0 + 0\right) = \frac{\pi}{8}$

• Calculate the volume V of the solid bounded by the three coordinate planes and the plane $2x + y + 4z = 4$.

To find the volume V of the solid, we use a double integral. The volume is given by: V = $\iint_R f(x, y) dA$ where f(x, y) represents the height of the solid above the region R in the xy-plane. The plane equation 2x +y +4z = 4 can be rearranged to express z in terms of x and y: z = $\frac{1}{4}(4-2x-y)$ (Figure iii).

Determining the Region R

The region R is the projection (or the “footprint”) of the solid onto the xy-plane (where $z = 0$). Setting $z=0$ gives the boundary line $2x + y = 4$, or $y = 4 - 2x$.

1. $x$ ranges from $0$ to the x-intercept at $x = 2$.
2. y ranges from 0 to $4 - 2x$.

Therefore, the region R, shown in Figure iii, is $R = \{(x, y): 0 \le x \le 2, 0 \le y \le 4 -2x \}$

1. Setting up the double integral. V = $\iint_R f(x, y)dA = \iint_R \frac{1}{4}(4-2x-y)dA = \int_{0}^{2} (\int_{0}^{4-2x} \frac{1}{4}(4-2x-y)dy)dx$
2. Computing the Inner Integral. $\int_{0}^{4-2x} \frac{1}{4}(4-2x-y)dy =$
   Factor out the constant: $\frac{1}{4} \int_{0}^{4-2x} (4-2x-y)dy = \frac{1}{4}[(4-2x)y-\frac{y^2}{2}]\bigg|_{0}^{4-2x}$
   $ = \frac{1}{4}[(4-2x)(4-2x)-\frac{(4-2x)^2}{2}-0] = \frac{1}{4}[(4-2x)^2-\frac{(4-2x)^2}{2}] = \frac{1}{4}[\frac{(4-2x)^2}{2}] = \frac{1}{8}(4-2x)^2.$
3. Computing the Outer Integral. $\int_{0}^{2} \frac{1}{8}(4-2x)^2 dx$
   Use u-substitution. Let $u = 4-2x \implies du = -2dx$. When $x = 0 \implies u = 4$. When $x = 2 \implies u = 0$.
   $\int_{4}^{0} \frac{1}{8}u^2\left(\frac{-1}{2}\right)du = -\frac{1}{16} \int_{4}^{0} u^2du$
   The negative sign allows us to flip the integration bounds back to the natural, standard order (that is, smaller to larger): $= \frac{1}{16} \int_{0}^{4} u^2du = \frac{1}{16} \left[\frac{u^3}{3}\right]_{0}^{4} = \frac{1}{16} \cdot \frac{64}{3} = \frac{4}{3}$

• Compute $\int_{0}^{1} (\int_{\sqrt{x}}^{x} \frac{e^y}{y}dy)dx$.

In its current form, this integral cannot be solved because $\frac{e^y}{y}$ lacks an elementary antiderivative. We must exchange the order of integration, which will structurally change the bounds of the integrals(Figure v).

Region of Integration.

To change the order of integration, we first need to understand the region of integration. The region is bounded by: x ranging from 0 to 1, y ranging from $\sqrt{x}$ to x (notice that the inner integral moves from a larger value to a smaller one because $\sqrt{x} \gt x$ on this interval).

When changing the order of integration, y will go from 0 to 1, and for a given y, x will range from $y$ to $y^2$. To maintain consistency with the original orientation, the limits integrate from the larger value $y$ to the smaller value $y^2$. Therefore, the new integral becomes: $\int_{0}^{1} (\int_{\sqrt{x}}^{x} \frac{e^y}{y}dy)dx = \int_{0}^{1} (\int_{y}^{y^2} \frac{e^y}{y}dx)dy$.

Compute the Inner Integral: $\int_{y}^{y^2} \frac{e^y}{y}dx = \frac{e^y}{y}x\bigg|_{y}^{y^2}=e^yy-e^y$

Compute the Outer Integral: $\int_{0}^{1} e^yy-e^ydy = \int_{0}^{1} e^yydy - \int_{0}^{1} e^ydy$

The second part requires Integration by Parts ($\int u dv = uv - \int v du$): Let $u = y$, $dv = e^y dy$. Then, $du = dy$, $v = e^y$.

$\int e^yydy = ye^y -\int e^ydy$

$ye^y\bigg|_{0}^{1} = e -0 = e$

$\int e^ydy = e^y\bigg|_{0}^{1} =e -1$

Putting it all together:

$\int_{0}^{1} e^yy-e^ydy = (e -(e-1))-(e -1)= e -e +1 -e +1 = 2 -e$

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

1. Changing the Order of Integration. Since the boundaries are constants, we can use either order.
   $\iint_R (2y-3x^2y^2) dA = \int_{0}^{1}\int_{0}^{2} (2y-3x^2y^2dy)dx = \int_{0}^{2}\int_{0}^{1} (2y-3x^2y^2dx)dy$
2. Compute the Inner Integral: $\int_{0}^{1} (2y-3x^2y^2dx) = 2yx-3y^2\frac{x^3}{3}=2yx-y^2x^3\bigg|_{0}^{1} = (2y·1-y^2·1^3)-(2y·0-y^2·0^3) = 2y -y^2$
3. Compute the Outer Integral: $\int_{0}^{2} (2y -y^2)dy=\frac{2y^2}{2}-\frac{y^3}{3} = y^2 -\frac{y^3}{3}\bigg|_{0}^{2} = 2^2-\frac{2^3}{3}=\frac{12}{3}-\frac{8}{3}=\frac{4}{3}$

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.