Double integrals II

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

Recall

• A function of two variables f: ℝ x ℝ → ℝ assigns to each ordered pair in its domain a unique real number, e.g., Area = $\frac{1}{2}b·h$, z = f(x, y) = 2x + 3y, f(x, y) = x² + y², e^x+y, etc.

• Partial derivatives are derivatives of a function of multiple variables, say f(x₁, x₂, ···, x_n) with respect to one of those variables, with the other variables held constant. They measure how the function changes as one variable changes, while keeping the others constant. $\frac{\partial f}{\partial x}(x_0, y_0) = \lim_{\Delta x \to 0} \frac{f(x_0+\Delta x, y_0)-f(x_0, y_0)}{\Delta x}$

Geometrically, the partial derivative $\frac{\partial f}{\partial x}(x_0, y_0)$ at a point (x₀, y₀) can be interpreted as the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane y = y₀. Similarly, $\frac{\partial f}{\partial y}(x_0, y_0)$ represents the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane x = x₀

• $\frac{dw}{ds}\bigg|_{\vec{u}}$ is the directional derivative in the direction of ${\vec{u}}$, the rate of change of the function w as we move in the direction of $\vec{u}$.

$\frac{dw}{ds}\bigg|_{\vec{u}} = ∇w·\vec{u} = |∇w|·|\vec{u}|cos(θ) = |∇w|·cos(θ)$ where θ is the angle between the gradient and the given unit vector.

• The directional derivative is maximized when cos(θ) = 1 ↭ θ = 0, that means, $\vec{u}$ is in the same direction as the gradient ∇w, $\vec{u}=dir(∇w).$

The gradient is the direction in which the function increases fastest (the direction of the steepest ascent) at a given point, and |∇w| = $\frac{dw}{ds}\bigg|_{\vec{u}=dir(∇w)}$.

The directional derivative is minimized when cos(θ) = -1 ↭ θ = 180 ↭ $\vec{u}$ is in the opposite direction of the gradient ∇w. Futhermore, $\frac{dw}{ds}\bigg|_{\vec{u}} = 0$ ↭ cos(θ) = 0 ↭ θ = 90° ↭ $\vec{u}$ ⊥ ∇w.

• The Chain Rule. Let w = f(x, y, z) where x = x(t), y = y(t), and z = z(t) are functions of another variable t, then the derivate of f with respect to t is given by $\frac{dw}{dt} = f_x\frac{dx}{dt} +f_y\frac{dy}{dt}+f_z\frac{dz}{dt}, \frac{dw}{dt} = w_x\frac{dx}{dt} +w_y\frac{dy}{dt}+w_z\frac{dz}{dt} = ∇w·\frac{d\vec{r}}{dt}~$ where ∇w = ⟨w_x, w_y, w_z⟩ and $\frac{d\vec{r}}{dt}=⟨\frac{d\vec{x}}{dt}, \frac{d\vec{y}}{dt}, \frac{d\vec{z}}{dt}⟩$

• The gradient vector is defined as follows: ∇w = ⟨w_x, w_y, w_z⟩ = $⟨\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z}⟩$. And $\frac{d\vec{r}}{dt}$ is the vector of derivatives of x, y, and z with respect to t: $\frac{d\vec{r}}{dt} = ⟨\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}⟩$.

The gradient vector is a vector that points in the direction of the steepest increase of the function at a given point. It is perpendicular (orthogonal) to the level surfaces of the function. This means that the gradient points in the direction of the steepest ascent or increase, and there is no change in the function’s value along the level surface, making the gradient perpendicular to it.

• Lagrange multipliers are a strategy for finding the local maxima and minima of a function, say f(x, y, z), where the variables x, y, and z are not independent, but subject to an equality constraint, g(x, y, z) = c. This method uses the fact that any point where f(x, y, z) reaches a constraint extremum (maximum or minimum), the gradient of f is parallel to the gradient of g. Formally, this relationship is expressed as ∇f = λ∇g where λ is the Lagrange multiplier.

• 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 (e.g. in the xy-plane). Instead of just finding or calculating an area, with a double integral, we’re looking for the volume under the surface (or below the graph) z = f(x, y). It is denoted or written as $\iint_R f(x, y) dA$

• Fubini’s Theorem. Suppose f(x, y) is a continuous function defined on a rectangular region R = [a, b] x [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: $\int \int_R f(x, y)dA = \int_{a}^{b}\int_{c}^{d} f(x, y)dydx = \int_{c}^{d}\int_{b}^{a} f(x, y)dxdy$

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, e.g., $\int_{0}^{1} (\int_{0}^{2} dx)dy = \int_{0}^{2} (\int_{0}^{1} dy)dx$ (Figure i), $\int_{0}^{2}(\int_{1}^{3} xy^2dy)dx = \int_{1}^{3} (\int_{0}^{2} xy^2dx)dy$.

This method allows us to simplify calculations by choosing the most convenient order of integration based on the given problem.

Solved exercises

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

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

🚀 Now, let’s handle sin(tan^-1(x)). Recall the trigonometric identity for sine of the arctangent function: sin(tan^-1(x)) = $\frac{x}{\sqrt{1+x^2}}$ (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 + x², du = 2xdx, $\int_{0}^{1} \frac{x}{\sqrt{1+x^2}}dx = \frac{1}{2} \int_{1}^{2}\frac{1}{\sqrt{u}}du = \frac{1}{2}2\sqrt{u} = \sqrt{1+x^2}$

$\int_{0}^{1} (\frac{1}{\sqrt{2}}-{\frac{x}{\sqrt{1+x^2}}})dx = (\frac{1}{\sqrt{2}}x-\sqrt{1+x^2})\bigg|_{0}^{1} = \frac{1}{\sqrt{2}}-\sqrt{2}-(0-1) = \frac{1}{\sqrt{2}}-\frac{2}{\sqrt{2}}+1 = 1 -\frac{1}{\sqrt{2}}$

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

Changed the order of integration to simplify the computation. Sometimes, changing the order of integration can simplify the computation significantly. The original double integral is set up with y ranging from 0 to 1 and x ranging from y to 1. To change the order of integration, consider the region of integration where x ranges from 0 to 1. Now, for each fixed x, y ranges from 0 to x (Figure D).

Thus, the double integral can be rewritten by swapping the order of integration:

$\int_{0}^{1} \int_{y}^{1}\frac{1}{1+x^4}dxdy = \int_{0}^{1} \int_{0}^{x}\frac{1}{1+x^4}dydx.$

Compute the inner integral. $\int_{0}^{x}\frac{1}{1+x^4}dy = \frac{1}{1+x^4}y\bigg|_{0}^{x} = \frac{x}{1+x^4}$.

Compute the outer integral. $\int_{0}^{1} \frac{x}{1+x^4}dx$ =[Substitution, u = x², du = 2xdx] $\frac{1}{2} \int_{0}^{1}\frac{du}{1+u^2} = \frac{1}{2}tan^{-1}(u)\bigg|_{0}^{1} =$

The integral $\int \frac{1}{1+u^2}du$ = tan^-1(u) is a well-known result.

$\frac{1}{2}(tan^{-1}(1)-tan^{-1}(0)) = \frac{1}{2}(\frac{π}{4}-0) = \frac{π}{8}$

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

Changed the order of integration to simplify the computation. Sometimes, changing the order of integration can simplify the computation significantly. The original double integral is set up with x ranging from 0 to 1 and y ranging from $\sqrt{x}$ to 1. To change the order of integration, consider the following region of integration: y will range from 0 to 1. For each fixed y, x will now range from 0 to y² (Figure 1).

Thus, the double integral can be rewritten by swapping the order of integration:

$\int_{0}^{1} \int_{\sqrt{x}}^{1}cos(y^3)dydx = \int_{0}^{1} \int_{0}^{y^2}cos(y^3)dxdy$

Compute the inner integral. $\int_{0}^{y^2}cos(y^3)dx = xcos(y^3)\bigg|_{0}^{y^2} = y^2cos(y^3)$.

Compute the outer integral. $\int_{0}^{1} y^2cos(y^3)dy$

This integral can be evaluated using a substitution method: $u = y^3, du = 3y^2dy ↭ \frac{du}{3} = y^2dy$. When y = 0, u = 0. When y = 1, u = 1. Thus, the integral becomes:

$\frac{1}{3}\int_{0}^{1} cos(u)du = \frac{1}{3} sin(u)\bigg|_{0}^{1} = \frac{1}{3}(sin(1)-sin(0))= \frac{1}{3}sin(1)$.

• Compute the double integral: $\int_R xye^{xy^2}dA$ on R = [0, 1] x [0, 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 = [a, b] x [c, d] 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:

$\int \int_R f(x, y)dA = \int_{a}^{b}\int_{c}^{d} f(x, y)dydx = \int_{c}^{d}\int_{b}^{a} f(x, y)dxdy$

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 and then with respect to x (the right double integral).

Compute the inner integral. $\int_{0}^{1} xye^{xy^2}dy =$[Substitution u = xy², du = 2xydy ⇒ $\frac{1}{2}du = xydy$] $\frac{1}{2}\int_{0}^{1} e^udu = \frac{e^u}{2} = \frac{1}{2}e^{xy^2}\bigg|_{0}^{1} = \frac{1}{2}(e^x-1)$

Compute the outer integral. $\int_{0}^{1} \frac{1}{2}(e^x-1)dx = \frac{1}{2}(e^x-x)\bigg|_{0}^{1} = \frac{1}{2}[(e-1)-(1-0)] = \frac{1}{2}(e-2)$

• 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 can simplify the computation significantly. The original double integral is set up with x ranging from 0 to 8 and y ranging from $\sqrt[3]{x}$ to 2. To change the order of integration, consider the region of integration: y will range from 0 to 2. For each fixed y, x will now range from 0 to y³ (Figure 2).

Thus, the double integral can be rewritten by swapping the order of integration:

$\int_{0}^{8} \int_{\sqrt[3]{x}}^{2}\frac{1}{y^4+1}dydx = \int_{0}^{2} \int_{0}^{y^3}\frac{1}{y^4+1}dxdy$

Compute the inner integral. $\int_{0}^{y^3}\frac{1}{y^4+1}dx = \frac{1}{y^4+1}x\bigg|_{0}^{y^3} = \frac{y^3}{y^4+1}$

Compute the outer integral. $\int_{0}^{2} \frac{y^3}{y^4+1} dy$.

This integral is a standard form, which can be solved using substitution. Let u = y⁴ + 1, so du = 4y³dy. Rewriting the integral in terms of u, we get:

$\int_{0}^{2} \frac{y^3}{y^4+1} dy = \frac{1}{4}\int \frac{1}{u}du = \frac{1}{4}ln|u| = \frac{1}{4}ln|y^4+1|\bigg|_{0}^{2} = \frac{1}{4}(ln(17)-ln(1)) = \frac{ln(17)}{4}$.

• Let z = f(x, y) = 1 -x² -y² and the region is not going to be determined by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, but by the region between the paraboloid and the xy plane, that is, looking at the part of the paraboloid that sits above the xy plane.

Therefore, R (figure D) is a quarter disk where x²+ y² ≤ 1, x ≥ 0, and y ≥ 0. It is basically the same integral, but the bounds change, meaning that for a particular x₀, y varies from 0 to $\sqrt{1-x_0^2}$

Calculation of the Double Integral: $\iint_R f(x, y) dA\ = \int_{0}^{1}(\int_{0}^{\sqrt{1-x^2}} 1-x^2-y^2dy)dx =$[🚀]

Compute the Inner Integral: (Integrate with respect to y, then evaluate the inner integral) $\int_{0}^{\sqrt{1-x^2}} 1-x^2-y^2dy = (y-x^2y-\frac{y^3}{3})\bigg|_{0}^{\sqrt{1-x^2}} = \sqrt{1-x^2}-x^2\sqrt{1-x^2}-\frac{1}{3}(1-x^2)^{\frac{3}{2}}=(1-x^2)\sqrt{1-x^2}-\frac{1}{3}(1-x^2)^{\frac{3}{2}} = (1-x^2)^1(1-x^2)^\frac{1}{2}-\frac{1}{3}(1-x^2)^{\frac{3}{2}} = (1-x^2)^{\frac{3}{2}}-\frac{1}{3}(1-x^2)^{\frac{3}{2}} = \frac{2}{3}(1-x^2)^{\frac{3}{2}}$

Compute the Outer Integral (Integrate with respect to x, then evaluate the outer integral): =[🚀] $\int_{0}^{1} \frac{2}{3}(1-x^2)^{\frac{3}{2}}dx =$[Substitute x = sin(θ), $(1-x^2)^{\frac{1}{2}} = cos(θ), dx = cos(θ)dθ, 0 ≤ x ≤ 1 ⇒ 0 ≤ θ ≤ \frac{π}{2}$] $\int_{0}^{\frac{π}{2}} \frac{2}{3}(1-sin^2(θ))^{\frac{3}{2}}cos(θ)dθ = \frac{2}{3}\int_{0}^{\frac{π}{2}}cos^3(θ)·cos(θ)dθ = \int_{0}^{\frac{π}{2}} \frac{2}{3}cos^4(θ)dθ$.

$\int_{0}^{\frac{π}{2}} \frac{2}{3}cos^4(θ)dθ =[cos^2(θ)=\frac{1+cos(2θ)}{2}, cos^4(θ)=(\frac{1+cos(2θ)}{2})^2 = \frac{1}{4}(1+2cos(2θ)+cos^2(2θ)).\text{ Besides, } cos^{2}(2θ)=\frac{1+cos(4θ)}{2}] = \frac{2}{3}\int_{0}^{\frac{π}{2}}\frac{1}{4}(1+2cos(2θ)+\frac{1+cos(4θ)}{2})=\frac{2}{3}\frac{1}{4}\int_{0}^{\frac{π}{2}}(1+2cos(2θ)+\frac{1}{2}+\frac{cos(4θ)}{2})dθ = \frac{1}{6}(\frac{3}{2}θ+sin(2θ)+\frac{1}{8}sin(4θ))\bigg|_{0}^{\frac{π}{2}}=$

=$\frac{1}{6}(\frac{3}{2}·\frac{π}{2}+0+0) = \frac{π}{8}$

• 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 of the solid onto the xy-plane. It is bounded by the lines where the planes intersects the z-and y axes:

1. x ranges from 0 to 2 (from the z-intercept)
2. y ranges from 0 to 4 -2x.

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

Setting up the double integral.

V = $\iint_R f(x, y)dA = \iint_R \frac{1}{4}(4-2x-y)dA = \int_{0}^{2} (\int_{0}^{4-2x} \frac{1}{4}(4-2x-y)dy)dx$

Computing the Inner Integral.

$\int_{0}^{4-2x} \frac{1}{4}(4-2x-y)dy =$[Factor out the constant] $\frac{1}{4} \int_{0}^{4-2x} (4-2x-y)dy = \frac{1}{4}[(4-2x)y-\frac{y^2}{2}]\bigg|_{0}^{4-2x} = \frac{1}{4}[(4-2x)(4-2x)-\frac{(4-2x)^2}{2}-0] = \frac{1}{4}[(4-2x)^2-\frac{(4-2x)^2}{2}] = \frac{1}{4}[\frac{(4-2x)^2}{2}] = \frac{1}{8}(4-2x)^2.$

Computing the Outer Integral

V = $\int_{0}^{2} (\int_{0}^{4-2x} \frac{1}{4}(4-2x-y)dy)dx = \int_{0}^{2} (\frac{1}{8}(4-2x)^2)dx = [Let, u = 4-2x⇒ du = -2dx, x = 0 ⇒u = 4; x = 2⇒u = 0] = \int_{4}^{0} \frac{1}{8}u^2(\frac{-1}{2}du) = \frac{-1}{16} \int_{4}^{0} u^2du = \frac{-1}{16} \frac{u^3}{3}\bigg|_{0}^{4} = \frac{-1}{16}·\frac{4^3}{3} = \frac{-1}{16}·\frac{64}{3} = \frac{-64}{48} = \frac{-2^6}{2^4·3}=\frac{-4}{3}.$

Since the volume cannot be negative, we take the positive value, $\frac{4}{3}.$

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

In its actual form, this integral is difficult to compute. We will exchange the order of integration, which will typically affect the bounds of the integrals and it is not necessarily easily seen (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 from 0 to 1, y from $\sqrt{x}$ to x. When changing the order of integration, y will go from 0 to 1, and for a given y, x will range from y to y². Therefore, the new integral will be:

$\int_{0}^{1} (\int_{\sqrt{x}}^{x} \frac{e^y}{y}dy)dx = \int_{0}^{1} (\int_{y}^{y^2} \frac{e^y}{y}dx)dy$.

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

$\int_{0}^{1} e^yy-e^ydy.$

This integral can be split into two parts: $\int_{0}^{1} e^yy.$

Integration by parts the first part of the integrand, ∫udv=uv−∫vdu where u = y, dv = e^ydy, 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)| 0 ≤ x ≤ 1, 0 ≤ y ≤ 2}

Changing the Order of Integration

$\iint_R (2y-3x^2y^2) dA = \int_{0}^{1}\int_{0}^{2} (2y-3x^2y^2dy)dx = \int_{0}^{2}\int_{0}^{1} (2y-3x^2y^2dx)dy$

Compute the Inner Integral

$\int_{0}^{1} (2y-3x^2y^2dx) = 2yx-3y^2\frac{x^3}{3}=2yx-y^2x^3\bigg|_{0}^{1} = (2y·1-y^2·1^3)-(2y·0-y^2·0^3) = 2y -y^2$

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.