]> Double integral

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 3: Many Variables; Section 2: Integration; page 1

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Double Integral

Definition

As we already know, the Cartesian plane consists of two perpendicular axes, denoted usually by x and y. In the case of one variable, x was used as the independent variable and y as the function. The definite integral of the function y with respect to x in the limits (a,b) was defined as

$\begin{array}{l} \int_{a}^{b} y (x) d x = \lim_{n \to \infty} \sum_{i = 1}^{n} y (x_{i}) Δ x_{i} \\ where all Δ x_{i}' s \to 0 \\ and x_{1} = a, x_{n} = b \end{array}}$	(3.2.1.1)

For a functions of two Cartesian variables f(x, y), the integration is defined in a region of the (x, y) plane, considering this way the integral as definite. Any two segments Δx and Δy correspond to a rectangular cell in this plane. The "area" of such a cell has the physical dimensions of the product ΔxΔy. By summing up all the elements,

$f (x, y) Δ x Δ y$	(3.2.1.2)

over the cells contained partially or fully in the region, and by doing the double limit of the segments Δx→0 and Δy→0 , we obtain the double integral of the function f in the region:

$\int \int f (x, y) d x d y = \lim_{\begin{matrix} Δ x \to 0 \\ Δ y \to 0 \end{matrix}} \sum f (x, y) Δ x Δ y$	(3.2.1.3)

One can restrict the requirement for the cells, to be only fully contained in the region, without affecting the outcome of the integral. Notice that any value the function f within the appropriate cell of (3.2.1.2) will do.

This definition for the simple case of f =1, is illustrated in Fig. Definition of double integral.

Evaluation

We evaluate a double integral, by iteration of two single integrals, first on x and then on y, or in reverse order. Since the evaluation of a single integral is a known procedure, we have only to learn how to define the limits of the iterated integrals. If the first iteration is along x, then the limits of the integral are y dependent, according to the boundary of the region of integration, and therefore the value of the first iteration is a function of y. The second iteration is a regular single integral, with limits of constant values.

This will be clarified for the integration over a convex region. A region is convex, if a straight segment connecting any two points (of the region), lies entirely in the region. For example a full circle is a convex region. But if some part is missing or added to, it may lose its convexity, pending on the shape of this part. This is illustrated in Fig. Convexity of a region.

For a given value of y = y_o in a convex region, the connecting segment between the minimal and the maximal x values is included in this region, and one can integrate along x, obtaining

$\int_{x_{\min} (y_{0})}^{x_{\max} (y_{0})} f (x, y_{0}) d x = I (y_{0})$	(3.2.1.4)

where I depends on the value of y_o. Since (3.2.1.4) exists for any y value in the region, one can rewrite it in a more general form

$\int_{x_{\min} (y)}^{x_{\max} (y)} f (x, y) d x = I (y)$	(3.2.1.5)

where the functions $x_{\min} (y)$ and $x_{\max} (y)$ represent the two branches of the region's boundary.

The double integral (3.2.1.3) can be rewritten now as

$\begin{array}{l} \int \int f (x, y) d y d x = \lim_{\begin{matrix} Δ x \to 0 \\ Δ y \to 0 \end{matrix}} \sum f (x, y) Δ y Δ x = \\ = \lim_{Δ y \to 0} \sum [Δ y \lim_{Δ x \to 0} \sum f (x, y) Δ x] = \\ = \lim_{Δ y \to 0} \sum [Δ y \int_{x_{\min} (y)}^{x_{\max} (y)} f (x, y) d x] = \\ = \lim_{Δ y \to 0} \sum [I (y) Δ y] = \int_{y_{\min}}^{y_{\max}} I (y) d y \end{array}}$	(3.2.1.6)

where the function I(y) from (3.2.1.5) was used for the second iteration. The limits of the second integration are the minimal and maximal values of y in the region. In order to summarize:

$\int \int f (x, y) d x d y = \int_{y_{\min}}^{y_{\max}} d y \int_{x_{\min} (y)}^{x_{\max} (y)} f (x, y) d x$	(3.2.1.7)

If the first integration would be done with respect to y, the limits are $y_{\min} (x)$ and $y_{\max} (x)$ , representing the two branches of the boundaries of the region, while the second iteration represents an integration with respect to x between the minimal and maximal x values of the region:

$\int \int f (x, y) d x d y = \int_{x_{\min}}^{x_{\max}} d x \int_{y_{\min} (x)}^{y_{\max} (x)} f (x, y) d y$	(3.2.1.8)

In both cases, (3.2.1.7) and (3.2.1.8), first one has to evaluate the inner-most integral. Only in the simplest case of a straight rectangular region

$\begin{array}{l} for {\begin{cases} x_{1} \leq x \leq x_{2} \\ y_{1} \leq y \leq y_{2} \end{cases} \\ \int \int f (x, y) d x d y = \int_{y_{1}}^{y_{2}} d y \int_{x_{1}}^{x_{2}} f (x, y) d x \end{array}}$	(3.2.1.9)

the limits of both integrations are constants.

The following example of a convex region

$\begin{array}{l} region's boundaries {\begin{cases} x = 0 (y axis) \\ y = 0 (x axis) \\ x + y = 1 \end{cases} \\ \int \int f (x, y) d x d y = {\begin{cases} \int_{0}^{1} d y \int_{0}^{1 - y} f (x, y) d x \\ \int_{0}^{1} d x \int_{0}^{1 - x} f (x, y) d y \end{cases} \end{array}}$	(3.2.1.10)

is illustrated in Fig. Convex region.

If the region for integration is not convex, it may happen that a straight segment of x or y, which is needed for the integration, is not entirely included in the region. In such a case, one has to exclude from the integration, the non-included segments according to the particularity of the region.

As an example, the non-convex region bounded by the two parabolas

$\begin{array}{l} y_{\min} = 2 x^{2} - 1 \\ y_{\max} = x^{2} \end{array}}$	(3.2.1.11)

intersecting at the points (x,y)=(±1,1) is non-convex. We'll discuss three ways of iterations. Fig. Non-convex region is given as an illustration, and it is advisable to keep it open in parallel, for the study of this example.

1.
From the observation of the region, if the first integration is done with respect to y for a given value of x, then the segment of the integration is contained fully in the region, and does not have any gaps. Therefore one can write

$\int \int f (x, y) d x d y = \int_{- 1}^{1} d x \int_{y_{\min}}^{y_{\max}} f (x, y) d y$	(3.2.1.12)

where $y_{\min} and y_{\max}$ are from (3.2.1.11) and the limits for the integration over x correspond to the minimal and maximal x values of the region, exactly like it is for a convex region.

2.
Also from the observation, if the first integration is over x, the behaviour depends on the sign of y :

For negative y, the integration over x should be done between a minimal and a maximal values of x, corresponding to the curve $y_{\min}$ (3.2.1.11) , namely $x_{\min} = - \sqrt{\frac{y + 1}{2}} and x_{\max} = \sqrt{\frac{y + 1}{2}}$ . The integration over y should be done for the negative values of y, namely between the limits −1 and 0 . All this yields part of the required integral:

$\begin{array}{l} I_{1} = \int_{- 1}^{0} d y \int_{x_{\min}}^{x_{\max}} f (x, y) d x \\ {\begin{cases} x_{\min} = - \sqrt{\frac{y + 1}{2}} \\ x_{\max} = \sqrt{\frac{y + 1}{2}} \end{cases} \end{array}}$	(3.2.1.13)

For positive y, the integration over x can be done only by two separate segments. For each segment, the limits of the integral correspond to the curves $y_{\min} and y_{\max}$ (3.2.1.11) . The integration over y should be done this time from 0 to 1. One obtains two new expressions I₂ and I₃ that should be added to I₁ in order to obtain the required integral:

$\begin{array}{l} I_{2} = \int_{0}^{1} d y \int_{x_{\min}}^{x_{\max}} f (x, y) d x \\ {\begin{cases} x_{\min} = - \sqrt{\frac{y + 1}{2}} \\ x_{\max} = - \sqrt{y} \end{cases} \end{array}}$	(3.2.1.14)

$\begin{array}{l} I_{3} = \int_{0}^{1} d y \int_{x_{\min}}^{x_{\max}} f (x, y) d x \\ {\begin{cases} x_{\min} = \sqrt{y} \\ x_{\max} = \sqrt{\frac{y + 1}{2}} \end{cases} \end{array}}$	(3.2.1.15)

$\int \int f (x, y) d x d y = I_{1} + I_{2} + I_{3}$	(3.2.1.16)

3.
We could also extend the region to a convex one for doing the integration there, and subtracting afterwards the integral of the extension. This can be done, only if the function is integrable in the extended region. It may sound complicated, but as we'll see, it is not, and is frequently used.

By defining a new curve

$y_{new} = 1$	(3.2.1.17)

we can define two convex regions between the boundaries

$\begin{array}{l} 1. y_{\min} and y_{new} \\ 2. y_{\max} and y_{new} \end{array}}$	(3.2.1.18)

The subtraction of the integral over the region 2, from that over the region 1 (3.2.1.18), will give the required integral.

Interpretations

Any double integral (3.2.1.3) with f(x,y)=1 represents the "area" of the appropriate region. We already used the integral of the function y(x), with respect to the variable x in order to express an enclosed area. These two ways of expressing an area are two different approaches that solve the same problem.

The area of the triangular region from (3.2.1.10) will be used as a simple example:

$Area = {\begin{cases} \int_{0}^{1} d y \int_{0}^{1 - y} d x = \int_{0}^{1} (1 - y) d y = (y - \frac{y^{2}}{2}) \|_{0}^{1} = \\ \int_{0}^{1} d x \int_{0}^{1 - x} d y = \int_{0}^{1} (1 - x) d x = (x - \frac{x^{2}}{2}) \|_{0}^{1} = \end{cases}} = \frac{1}{2}$	(3.2.1.19)

As another example, the area of an ellipse

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$	(3.2.1.20)

will be calculated. This was already done by the use of a single integral in an exercise from the previous chapter, giving the area $π a b$ . By the use of a double double integral, one obtains the area A

${\begin{cases} A = \int_{x_{1}}^{x_{2}} d x \int_{y_{1} (x)}^{y_{2} (x)} d y = \int_{- a}^{a} d x \int_{- \frac{b}{a} \sqrt{a^{2} - x^{2}}}^{\frac{b}{a} \sqrt{a^{2} - x^{2}}} d y = \frac{2 b}{a} \int_{- a}^{a} d x \sqrt{a^{2} - x^{2}} \\ (substitution: x = asin u, d x = acos u d u, \sqrt{a^{2} - x^{2}} = acos u) \\ A = 2 a b \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} u d u = 2 a b \frac{π}{2} = π a b \end{cases}}$	(3.2.1.21)

{\begin{cases} A = \int_{x_{1}}^{x_{2}} d x \int_{y_{1} (x)}^{y_{2} (x)} d y = \int_{- a}^{a} d x \int_{- \frac{b}{a} \sqrt{a^{2} - x^{2}}}^{\frac{b}{a} \sqrt{a^{2} - x^{2}}} d y = \frac{2 b}{a} \int_{- a}^{a} d x \sqrt{a^{2} - x^{2}} \\ (substitution: x = asin u, d x = acos u d u, \sqrt{a^{2} - x^{2}} = acos u) \\ A = 2 a b \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} u ​ d u = 2 a b \frac{π}{2} = π a b \end{cases}}

(3.2.1.21)

as expected.

We know from chapter 2 that the single integral of a curve y(x) over an interval of a variable x, represents the "area" between the curve and the x axis in the interval. By the same reasoning, a double integral of a surface z(x,y) over a (x,y) region of the variables, represents the "volume" under this surface.

As an example of the volume under a surface, we'll use the hyperbolic paraboloid, studied in the previous section.

$\begin{array}{l} z (x, y) = \frac{x y}{2} + 5 \\ in the region {\begin{cases} - 2 \leq x \leq 2 \\ - 2 \leq y \leq 2 \end{cases} \end{array}}$	(3.2.1.22)

which for convenience is translated in the z direction by 5 units, and for simplicity the region is a square. The example is illustrated in Fig. Volume under a surface.

The volume V in this example is

$\begin{array}{l} V = \int_{- 2}^{2} d y \int_{- 2}^{2} d x (\frac{x y}{2} + 5) = \int_{- 2}^{2} d y (y \frac{x^{2}}{4} + 5 x) \|_{- 2}^{2} = \\ = \int_{- 2}^{2} d y (y + 10 - y + 10) = 20 \int_{- 2}^{2} d y = 80 \end{array}}$	(3.2.1.23)

As another example, we'll calculate the volume of the ellipsoid

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1$	(3.2.1.24)

Half of its volume is enclosed between the surface corresponding to positive z

$z = c \sqrt{1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}}$	(3.2.1.25)

and the (x,y) plane (corresponding to z=0), therefore the volume V can be calculated over the same region of coordinates as in (3.2.1.21),

${\begin{cases} \frac{V}{2} = \int_{x_{1}}^{x_{2}} d x \int_{y_{1} (x)}^{y_{2} (x)} z (x, y) d y = c \int_{- a}^{a} d x \int_{- \frac{b}{a} \sqrt{a^{2} - x^{2}}}^{\frac{b}{a} \sqrt{a^{2} - x^{2}}} d y \sqrt{1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}} \\ (substitution: \sin u = \frac{\frac{y}{b}}{\sqrt{1 - \frac{x^{2}}{a^{2}}}}, d y = b \sqrt{1 - \frac{x^{2}}{a^{2}}} \cos u d u) \\ \frac{V}{2} = c b \int_{- a}^{a} d x (1 - \frac{x^{2}}{a^{2}}) \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} u d u = \frac{c b π}{2} \int_{- a}^{a} d x (1 - \frac{x^{2}}{a^{2}}) = \\ = \frac{π b c}{2} (x - \frac{x^{3}}{3 a^{2}}) \|_{- a}^{a} = \frac{π b c}{2} \frac{4 a}{3} \Rightarrow V = \frac{4 π}{3} a b c \end{cases}}$	(3.2.1.26)

{\begin{cases} \frac{V}{2} = \int_{x_{1}}^{x_{2}} d x \int_{y_{1} (x)}^{y_{2} (x)} z (x, y) d y = c \int_{- a}^{a} d x \int_{- \frac{b}{a} \sqrt{a^{2} - x^{2}}}^{\frac{b}{a} \sqrt{a^{2} - x^{2}}} d y \sqrt{1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}} \\ (substitution: \sin u = \frac{\frac{y}{b}}{\sqrt{1 - \frac{x^{2}}{a^{2}}}}, d y = b \sqrt{1 - \frac{x^{2}}{a^{2}}} \cos u d u) \\ \frac{V}{2} = c b \int_{- a}^{a} d x (1 - \frac{x^{2}}{a^{2}}) \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} u d u = \frac{c b π}{2} \int_{- a}^{a} d x (1 - \frac{x^{2}}{a^{2}}) = \\ = \frac{π b c}{2} (x - \frac{x^{3}}{3 a^{2}}) |_{- a}^{a} = \frac{π b c}{2} \frac{4 a}{3} \Rightarrow V = \frac{4 π}{3} a b c \end{cases}}

(3.2.1.26)

For the function of a single variable y(x), we defined its mean value in an interval (a,b) of x as the expression:

${〈 y 〉}_{(a, b)} = \frac{\int_{a}^{b} y (x) d x}{\int_{a}^{b} d x}$	(3.2.1.27)

In the case of a function of two variables $f (x, y)$ , one defines its mean value as

$〈 f 〉 = \frac{\int \int f (x, y) d x d y}{\int \int d x d y}$	(3.2.1.28)

in the region of integration.

As an example, the mean value z(x,y) (3.2.1.22-23) in the specified region is

$〈 z 〉 = \frac{\int \int z d x d y}{\int \int d x d y} = \frac{80}{16} = 5$	(3.2.1.29)

This means that

$〈 z 〉 = 〈 \frac{x y}{2} + 5 〉 = 5 \Rightarrow 〈 \frac{x y}{2} 〉 = 0$	(3.2.1.30)

or that $\frac{x y}{2}$ has balanced positive and negative values in the region.

Centre of mass

Centre of mass (or centre of gravitation) is an important notion of physics, which was obtained in equation (3.1.5.16) as the solution of an example. The definition is: For n physical points with masses m_i located in space with fixed coordinates (x, y, z)_i=(x_i, y_i, z_i), where i=1,2,....,n, the centre of mass is located at

$x_{c m} = \frac{\sum_{i = 1}^{n} m_{i} x_{i}}{\sum_{i = 1}^{n} m_{i}}, y_{c m} = \frac{\sum_{i = 1}^{n} m_{i} y_{i}}{\sum_{i = 1}^{n} m_{i}}, z_{c m} = \frac{\sum_{i = 1}^{n} m_{i} z_{i}}{\sum_{i = 1}^{n} m_{i}}$	(3.2.1.31)

In the case of a solid body, the sums of (3.2.1.31) become integrals. For example in the case of a two dimensional planar body in a region of (x,y), with a density of σ(x,y) mass/area, one obtains:

$\begin{array}{l} x_{c m} = \frac{\int x d m}{\int d m} = \frac{\int x d m}{M} = \frac{\int \int x σ (x, y) d x d y}{\int \int σ (x, y) d x d y} \\ y_{c m} = \frac{\int y d m}{\int d m} = \frac{\int y d m}{M} = \frac{\int \int y σ (x, y) d x d y}{\int \int σ (x, y) d x d y} \end{array}}$	(3.2.1.32)

where M is the total mass.

As an example we'll use a planar body in the triangular region of (3.2.1.10), with a mass density decreasing with x:

$\begin{array}{l} σ (x, y) = a (1 - x) \\ a = constant > 0 \end{array}}$	(3.2.1.33)

According to the limits of the integral (3.1.1.19), the denominator of (3.2.1.32) becomes

${\begin{cases} \int \int σ (x, y) d x d y = a \int_{0}^{1} d y \int_{0}^{1 - y} (1 - x) d x = \\ = a \int_{0}^{1} d y [x (1 - \frac{x}{2})] \|_{0}^{1 - y} = a \int_{0}^{1} d y [(1 - y) (\frac{2 - (1 - y)}{2})] = \\ = \frac{a}{2} \int_{0}^{1} (1 - y^{2}) d y = \frac{a}{2} (y - \frac{y^{3}}{3}) \|_{0}^{1} = \frac{a}{3} \end{cases}}$	(3.2.1.34)

and the centre of mass point is at

$\begin{array}{l} {\begin{cases} x_{c m} = \frac{3 a}{a} \int_{0}^{1} d y \int_{0}^{1 - y} (1 - x) x d x = 3 \int_{0}^{1} d y [x^{2} (\frac{1}{2} - \frac{x}{3})] \|_{0}^{1 - y} = \\ = \frac{3}{6} \int_{0}^{1} d y {(1 - y)}^{2} [3 - 2 (1 - y)] = \frac{1}{2} \int_{0}^{1} u^{2} (3 - 2 u) d u = \\ = \frac{1}{2} u^{3} (\frac{3}{3} - \frac{2 u}{4}) \|_{0}^{1} = \frac{1}{4} (u = 1 - y) \end{cases} \\ {\begin{cases} y_{c m} = 3 \int_{0}^{1} y d y \int_{0}^{1 - y} (1 - x) d x = \frac{3}{2} \int_{0}^{1} y (1 - y^{2}) d y = \\ = \frac{3}{2} (\frac{1}{2} - \frac{1}{4}) = \frac{3}{8} (\int_{0}^{1 - y} (1 - x) d x : see (3 .2 .1 .34)) \end{cases} \end{array}}$	(3.2.1.35)

\begin{array}{l} {\begin{cases} x_{c m} = \frac{3 a}{a} \int_{0}^{1} d y \int_{0}^{1 - y} (1 - x) x d x = 3 \int_{0}^{1} d y [x^{2} (\frac{1}{2} - \frac{x}{3})] |_{0}^{1 - y} = \\ = \frac{3}{6} \int_{0}^{1} d y {(1 - y)}^{2} [3 - 2 (1 - y)] = \frac{1}{2} \int_{0}^{1} u^{2} (3 - 2 u) d u = \\ = \frac{1}{2} u^{3} (\frac{3}{3} - \frac{2 u}{4}) |_{0}^{1} = \frac{1}{4} (u = 1 - y) \end{cases} \\ {\begin{cases} y_{c m} = 3 \int_{0}^{1} y d y \int_{0}^{1 - y} (1 - x) d x = \frac{3}{2} \int_{0}^{1} y (1 - y^{2}) d y = \\ = \frac{3}{2} (\frac{1}{2} - \frac{1}{4}) = \frac{3}{8} (\int_{0}^{1 - y} (1 - x) d x : see (3 .2 .1 .34)) \end{cases} \end{array}}

(3.2.1.35)

In the case of a uniform distribution of mass (constant density), one obtains

$x_{c m} = < x >, y_{c m} = < y >$	(3.2.1.36)

The point of a region corresponding to the mean value of the coordinates is called a centroid, therefore in the case of a constant density, the centre of mass of a body is equivalent to its centroid.

In the previous example (3.2.1.33), but with uniform mass distribution, we obtain

$\begin{array}{l} \int \int d x d y = \frac{1}{2} according to equation (3 .2 .1 .19) \\ x_{c m} = 2 \int_{0}^{1} d y \int_{0}^{1 - y} x d x = 2 \int_{0}^{1} d y \frac{{(1 - y)}^{2}}{2} = y (1 - y + \frac{y^{2}}{3}) \|_{0}^{1} = \frac{1}{3} \\ y_{c m} = 2 \int_{0}^{1} y d y \int_{0}^{1 - y} d x = 2 \int_{0}^{1} y (1 - y) d y = 2 (\frac{y^{2}}{2} - \frac{y^{3}}{3}) \|_{0}^{1} = \frac{1}{3} \end{array}}$	(3.2.1.37)

\begin{array}{l} \int \int d x d y = \frac{1}{2} according to equation (3 .2 .1 .19) \\ x_{c m} = 2 \int_{0}^{1} d y \int_{0}^{1 - y} x d x = 2 \int_{0}^{1} d y \frac{{(1 - y)}^{2}}{2} = y (1 - y + \frac{y^{2}}{3}) |_{0}^{1} = \frac{1}{3} \\ y_{c m} = 2 \int_{0}^{1} y d y \int_{0}^{1 - y} d x = 2 \int_{0}^{1} y (1 - y) d y = 2 (\frac{y^{2}}{2} - \frac{y^{3}}{3}) |_{0}^{1} = \frac{1}{3} \end{array}}

(3.2.1.37)

The equality x_cm=y_cm in (3.2.1.37) is due to the symmetry of the region under the (x,y) inversion.

Exercises

Reminder from chapter 2. The user can freely help himself with a table of integrals, for solving exercises.

Exercise 1. Two parabolas

y_{1} = x^{2} and y_{2} = a x^{2} - 1

where a is a constant, are enclosing an area in the (x,y) plane.

What is the possible range of the a values, and where do the parabolas intersect?
Use a double integral to calculate the area, by integrating first over y !
Use a double integral to calculate the area, by integrating first over x !
Use a double integral to calculate the area, by subtracting integrals over convex regions!

Exercise 2. Use a double integral to calculate the volume enclosed between the (x,y), (y,z), (x,z) planes, and the plane intersecting the axes at x=a , y=b , and z=c !

Exercise 3.

Use a double integral to calculate the volume enclosed between the partial surface of the elliptic paraboloid $z = h (1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}) \geq 0$ and the (x,y) plane, where h is a positive constant!
Express h in terms of c, if this volume equals the volume enclosed between the partial surface of the ellipsoid $z = c \sqrt{1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}} \geq 0$ and the (x,y) plane, while a and b keep the same values!

Exercise 4. The density of a flat solid body is uniformly distributed. The body consisits of two parts:

a rectangle with its base at the x axis between $- r \leq x \leq r$ , and with a height of y=h>0.
at $y \geq h$ , half of a circle with radius r and with centre at (x,y)=(0,h).

Make a sketch of the body!
What are the coordinates of the centre of mass?
What should be the height h (expressed in terms of r), if the centre of mass of the body is at the centre of the circle?
In such a case, what is the ratio of the masses of the rectangle versus that of the half circle?

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Math Animated™ Mathematical Introduction for Physics and Engineering by Samuel Dagan (Copyright © 2007-2020)

Chapter 3: Many Variables; Section 2: Integration; page 1

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Double Integral

Definition

Evaluation

Interpretations

Centre of mass

Exercises

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Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)