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Mathematics LibreTexts

14.2bE: Double Integrals Part 2 (Exercises)

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    13698
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    1) The region \(D\) bounded by \(y = x^3, \space y = x^3 + 1, \space x = 0,\) and \(x = 1\) as given in the following figure.

    A region is bounded by y = 1 + x cubed, y = x cubed, x = 0, and x = 1.

    a. Classify this region as vertically simple (Type I) or horizontally simple (Type II).

    Type:
    Type I but not Type II

    b. Find the area of the region \(D\).

    c. Find the average value of the function \(f(x,y) = 3xy\) on the region graphed in the previous exercise.

    Answer:
    \(\frac{27}{20}\)

     

    2) The region \(D\) bounded by \(y = \sin x, \space y = 1 + \sin x, \space x = 0\), and \(x = \frac{\pi}{2}\) as given in the following figure.

    A region is bounded by y = 1 + sin x, y = sin x, x = 0, and x = pi/2.

    a. Classify this region as vertically simple (Type I) or horizontally simple (Type II).

    Type:
    Type I but not Type II

    b. Find the area of the region \(D\).

    Answer:
    \(\frac{\pi}{2}\, \text{units}^2\)

    c. Find the average value of the function \(f(x,y) = \cos x\) on the region \(D\).

     

    3) The region \(D\) bounded by \(x = y^2 - 1\) and \(x = \sqrt{1 - y^2}\) as given in the following figure.

    A region is bounded by x = negative 1 + y squared and x = the square root of the quantity (1 minus y squared).

    a. Classify this region as vertically simple (Type I) or horizontally simple (Type II).

    Type:
    Type II but not Type I

    b. Find the volume of the solid under the graph of the function \(f(x,y) = xy + 1\) and above the region \(D\).

    Answer:
    \(\frac{1}{6}(8 + 3\pi)\, \text{units}^3\)

     

    4) The region \(D\) bounded by \(y = 0, \space x = -10 + y,\) and \(x = 10 - y\) as given in the following figure.

    A region is bounded by x = negative 10 + y, x = 10 minus y, and y = 0.

    a. Classify this region as vertically simple (Type I) or horizontally simple (Type II).

    Type:
    Type II but not Type I

    b. Find the volume of the solid under the graph of the function \(f(x,y) = x + y\) and above the region in the figure from the previous exercise.

    Answer:
    \(\frac{1000}{3}\, \text{units}^3\)

     

    5) The region \(D\) bounded by \(y = 0, \space x = y - 1, \space x = \frac{\pi}{2}\) as given in the following figure.

    A region is bounded by x = pi/2, y = 0, and x = negative 1 + y.

    Classify this region as vertically simple (Type I) or horizontally simple (Type II).

    Type:
    Type I and Type II

    6) The region \(D\) bounded by \(y = 0\) and \(y = x^2 - 1\) as given in the following figure.

    A region is bounded by y = 0 and y = negative 1 + x squared.

    Classify this region as vertically simple (Type I) or horizontally simple (Type II).

    Type:
    Type I and Type II

     

    7) Let \(D\) be the region bounded by the curves of equations \(y = cos \space x\) and \(y = 4 - x^2\) and the \(x\)-axis. Explain why \(D\) is neither of Type I nor II.

    Answer:
    The region \(D\) is not of Type I: it does not lie between two vertical lines and the graphs of two continuous functions \(g_1(x)\) and \(g_2(x)\). The region is not of Type II: it does not lie between two horizontal lines and the graphs of two continuous functions \(h_1(y)\) and \(h_2(y)\).

    8) Let \(D\) be the region bounded by the curves of equations \(y = x, \space y = -x\) and \(y = 2 - x^2\). Explain why \(D\) is neither of Type I nor II.

     

    In exercises 9 - 14, evaluate the double integral \(\displaystyle \iint_D f(x,y) \,dA\) over the region \(D\).

    9)  \(f(x,y) = 1\) and

    \(D = \big\{(x,y)| \, 0 \leq x \leq \frac{\pi}{2}, \space \sin x \leq y \leq 1 + \sin x \big\}\)

    Answer:
    \(\frac{\pi}{2}\)

    10)  \(f(x,y) = 2\) and

    \(D = \big\{(x,y)| \, 0 \leq y \leq 1, \space y - 1 \leq x \leq \arccos y \big\}\)

    11)  \(f(x,y) = xy\) and

    \(D = \big\{(x,y)| \, -1 \leq y \leq 1, \space y^2 - 1 \leq x \leq \sqrt{1 - y^2} \big\}\)

    Answer:
    \(0\)

    12)  \(f(x,y) = sin \space y\) and \(D\) is the triangular region with vertices \((0,0), \space (0,3)\), and \((3,0)\)

    13)  \(f(x,y) = -x + 1\) and \(D\) is the triangular region with vertices \((0,0), \space (0,2)\), and \((2,2)\)

    Answer:
    \(\frac{2}{3}\)

    14)  \(f(x,y) = 2x + 4y\) and

    \(D = \big\{(x,y)|\, 0 \leq x \leq 1, \space x^3 \leq y \leq x^3 + 1 \big\}\)

     

    In exercises 15 - 20, evaluate the iterated integrals.

    15)  \(\displaystyle \int_0^1 \int_{2\sqrt{x}}^{2\sqrt{x}+1} (xy + 1) \,dy \space dx\)

    Answer:
    \(\frac{41}{20}\)

    16)  \(\displaystyle \int_0^3 \int_{2x}^{3x} (x + y^2) \,dy \space dx\)

    17)  \(\displaystyle \int_1^2 \int_{-u^2-1}^{-u} (8 uv) \,dv \space du\)

    Answer:
    \(-63\)

    18)  \(\displaystyle \int_e^{e^2} \int_{\ln u}^2 (v + \ln u) \,dv \space du\)

    19)  \(\displaystyle \int_0^1 \int_{-\sqrt{1-4y^2}}^{\sqrt{1-4y^2}} 4 \,dx \space dy\)

    Answer:
    \(\pi\)

    20)  \(\displaystyle \int_0^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (2x + 4y^3) \,dx \space dy\)

     

    21)  Let \(D\) be the region bounded by \(y = 1 - x^2, \space y = 4 - x^2\), and the \(x\)- and \(y\)-axes.

    a. Show that

    \[\iint_D x\,dA = \int_0^1 \int_{1-x^2}^{4-x^2} x \space dy \space dx + \int_1^2 \int_0^{4-x^2} x \space dy \space dx\] by dividing the region \(D\) into two regions of Type I.

    b. Evaluate the integral \[\iint_D s \space dA.\]

    22)  Let \(D\) be the region bounded by \(y = 1, \space y = x, \space y = ln \space x\), and the \(x\)-axis.

    a. Show that

    \[\iint_D y^2 dA = \int_{-1}^0 \int_{-x}^{2-x^2} y^2 dy \space dx + \int_0^1 \int_x^{2-x^2} y^2 dy \space dx\] by dividing the region \(D\) into two regions of Type I, where \(D = \big\{(x,y)\,|\,y \geq x, y \geq -x, \space y \leq 2-x^2\big\}\).

    b. Evaluate the integral \[\iint_D y^2 dA.\]

    23)  Let \(D\) be the region bounded by \(y = x^2\), \(y = x + 2\), and \(y = -x\).

    a. Show that \[\iint_D x \space dA = \int_0^1 \int_{-y}^{\sqrt{y}} x \space dx \space dy + \int_1^2 \int_{y-2}^{\sqrt{y}} x \space dx \space dy\] by dividing the region \(D\) into two regions of Type II, where \(D = \big\{(x,y)\,|\,y \geq x^2, \space y \geq -x, \space y \leq x + 2\big\}\).

    b. Evaluate the integral \[\iint_D x \space dA.\]

    Answer:
    a. Answers may vary;
    b. \(\frac{8}{12}\)

     

    24)  The region \(D\) bounded by \(x = 0, y = x^5 + 1\), and \(y = 3 - x^2\) is shown in the following figure. Find the area \(A(D)\) of the region \(D\).

    A region is bounded by y = 1 + x to the fifth power, y = 3 minus x squared, and x = 0.

    25)  The region \(D\) bounded by \(y = cos \space x, \space y = 4 \space cos \space x\), and \(x = \pm \frac{\pi}{3}\) is shown in the following figure. Find the area \(A(D)\) of the region \(D\).

    A region is bounded by y = cos x, y = 4 + cos x, x = negative 1, and x = 1.

    Answer:
    \(\frac{8\pi}{3}\)

    26)  Find the area \(A(D)\) of the region \(D = \big\{(x,y)| \, y \geq 1 - x^2, y \leq 4 - x^2, \space y \geq 0, \space x \geq 0 \big\}\).

    27)  Let \(D\) be the region bounded by \( y = 1, \space y = x, \space y = ln \space x\), and the \(x\)-axis. Find the area \(A(D)\) of the region \(D\).

    Answer:
    \(\left(e - \frac{3}{2}\right)\, \text{units}^2\)

     

    28)  Find the average value of the function \(f(x,y) = sin \space y\) on the triangular region with vertices \((0,0), \space (0,3)\), and \((3,0)\).

    29)  Find the average value of the function \(f(x,y) = -x + 1\) on the triangular region with vertices \((0,0), \space (0,2)\), and \((2,2)\).

    Answer:
    \(\frac{2}{3}\)

     

    In exercises 30 - 33, change the order of integration and evaluate the integral.

    30)  \[\int_{-1}^{\pi/2} \int_0^{x+1} sin \space x \space dy \space dx\]

    31)  \[\int_0^1 \int_{x-1}^{1-x} x \space dy \space dx\]

    Answer:
    \[\int_0^1 \int_{x-1}^{1-x} x \space dy \space dx = \int_{-1}^0 \int_0^{y+1} x \space dx \space dy + \int_0^1 \int_-^{1-y} x \space dx \space dy = \frac{1}{3}\]

    32)  \[\int_{-1}^0 \int_{-\sqrt{y+1}}^{\sqrt{y+1}} y^2 dx \space dy\]

    33)  \[\int_{-1/2}^{1/2} \int_{-\sqrt{y^2+1}}^{\sqrt{y^2+1}} y \space dx \space dy\]

    Answer:
    \[\int_{-1/2}^{1/2} \int_{-\sqrt{y^2+1}}^{\sqrt{y^2+1}} y \space dx \space dy = \int_1^2 \int_{-\sqrt{x^2-1}}^{\sqrt{x^2-1}} y \space dy \space dx = 0\]

     

    34)  The region \(D\) is shown in the following figure. Evaluate the double integral \(\displaystyle \iint_D (x^2 + y) \,dA\) by using the easier order of integration.

    A region is bounded by y = negative 4 + x squared and y = 4 minus x squared.

    35)  The region \(D\) is shown in the following figure. Evaluate the double integral \(\displaystyle \iint_D (x^2 - y^2) \,dA\) by using the easier order of integration.

    A region is bounded by y to the fourth power = 1 minus x and y to the fourth power = 1 + x.

    Answer:
    \[\iint_D (x^2 - y^2) dA = \int_{-1}^1 \int_{y^4-1}^{1-y^4} (x^2 - y^2)dx \space dy = \frac{464}{4095}\]

    36)  Find the volume of the solid under the surface \(z = 2x + y^2\) and above the region bounded by \(y = x^5\) and \(y = x\).

    37)  Find the volume of the solid under the plane \(z = 3x + y\) and above the region determined by \(y = x^7\) and \(y = x\).

    Answer:
    \(\frac{4}{5}\, \text{units}^3\)

    38)  Find the volume of the solid under the plane \(z = 3x + y\) and above the region bounded by \(x = tan \space y, \space x = -tan \space y\), and \(x = 1\).

    39)  Find the volume of the solid under the surface \(z = x^3\) and above the plane region bounded by \(x = sin \space y, \space x = -sin \space y\), and \(x = 1\).

    Answer:
    \(\frac{5\pi}{32}\, \text{units}^3\)

    40)  Let \(g\) be a positive, increasing, and differentiable function on the interval \([a,b]\). Show that the volume of the solid under the surface \(z = g'(x)\) and above the region bounded by \(y = 0, \space y = g(x), \space x = a\), and \(x = b\) is given by \(\frac{1}{2}(g^2 (b) - g^2 (a))\).

    41)  Let \(g\) be a positive, increasing, and differentiable function on the interval \([a,b]\) and let \(k\) be a positive real number. Show that the volume of the solid under the surface \(z = g'(x)\) and above the region bounded by \(y = g(x), \space y = g(x) + k, \space x = a\), and \(x = b\) is given by \(k(g(b) - g(a)).\)

    42)  Find the volume of the solid situated in the first octant and determined by the planes \(z = 2\), \(z = 0, \space x + y = 1, \space x = 0\), and \(y = 0\).

    43)  Find the volume of the solid situated in the first octant and bounded by the planes \(x + 2y = 1\), \(x = 0, \space z = 4\), and \(z = 0\).

    Answer:
    \(1\, \text{units}^3\)

    44)  Find the volume of the solid bounded by the planes \(x + y = 1, \space x - y = 1, \space x = 0, \space z = 0\), and \(z = 10\).

    45)  Find the volume of the solid bounded by the planes \(x + y = 1, \space x - y = 1, \space x - y = -1, \space z = 1\), and \(z = 0\)

    Answer:
    \(2\, \text{units}^3\)

    46)  Let \(S_1\) and \(S_2\) be the solids situated in the first octant under the planes \(x + y + z = 1\) and \(x + y + 2z = 1\) respectively, and let \(S\) be the solid situated between \(S_1, \space S_2, \space x = 0\), and \(y = 0\).

    1. Find the volume of the solid \(S_1\).
    2. Find the volume of the solid \(S_2\).
    3. Find the volume of the solid \(S\) by subtracting the volumes of the solids \(S_1\) and \(S_2\).

    47)  Let \(S_1\) and \(S_2\) be the solids situated in the first octant under the planes \(2x + 2y + z = 2\) and \(x + y + z = 1\) respectively, and let \(S\) be the solid situated between \(S_1, \space S_2, \space x = 0\), and \(y = 0\).

    1. Find the volume of the solid \(S_1\).
    2. Find the volume of the solid \(S_2\).
    3. Find the volume of the solid \(S\) by subtracting the volumes of the solids \(S_1\) and \(S_2\).
    Answer:
    a. \(\frac{1}{3}\, \text{units}^3\)
    b. \(\frac{1}{6}\, \text{units}^3\)
    c. \(\frac{1}{6}\, \text{units}^3\)

    48)  Let \(S_1\) and \(S_2\) be the solids situated in the first octant under the plane \(x + y + z = 2\) and under the sphere \(x^2 + y^2 + z^2 = 4\), respectively. If the volume of the solid \(S_2\) is \(\frac{4\pi}{3}\) determine the volume of the solid \(S\) situated between \(S_1\) and \(S_2\) by subtracting the volumes of these solids.

    49)  Let \(S_1\) and \(S_2\) be the solids situated in the first octant under the plane \(x + y + z = 2\) and under the sphere \(x^2 + y^2 = 4\), respectively.

    1. Find the volume of the solid \(S_1\).
    2. Find the volume of the solid \(S_2\).
    3. Find the volume of the solid \(S\) situated between \(S_1\) and \(S_2\) by subtracting the volumes of the solids \(S_1\) and \(S_2\).
    Answer:
    a. \(\frac{4}{3}\, \text{units}^3\)
    b. \(2\pi\, \text{units}^3\)
    c. \(\frac{6\pi - 4}{3}\, \text{units}^3\)

    50)  [T] The following figure shows the region \(D\) bounded by the curves \(y = sin \space x, \space x = 0\), and \(y = x^4\). Use a graphing calculator or CAS to find the \(x\)-coordinates of the intersection points of the curves and to determine the area of the region \(D\). Round your answers to six decimal places.

    A region is bounded by y = sin x, y = x to the fourth power, and x = 0.

    51)  [T] The region \(D\) bounded by the curves \(y = cos \space x, \space x = 0\), and \(y = x^3\) is shown in the following figure. Use a graphing calculator or CAS to find the x-coordinates of the intersection points of the curves and to determine the area of the region \(D\). Round your answers to six decimal places.

    A region is bounded by y = cos x, y = x cubed, and x = 0.

    Answer:
    0 and 0.865474; \(A(D) = 0.621135\, \text{units}^3\)

    52)  Suppose that \((X,Y)\) is the outcome of an experiment that must occur in a particular region \(S\) in the \(xy\)-plane. In this context, the region \(S\) is called the sample space of the experiment and \(X\) and \(Y\) are random variables. If \(D\) is a region included in \(S\), then the probability of \((X,Y)\) being in \(D\) is defined as \(P[(X,Y) \in D] = \iint_D p(x,y)dx \space dy\), where \(p(x,y)\) is the joint probability density of the experiment. Here, \(p(x,y)\) is a nonnegative function for which \(\iint_S p(x,y) dx \space dy = 1\). Assume that a point \((X,Y)\) is chosen arbitrarily in the square \([0,3] \times [0,3]\) with the probability density

    \[p(x,y) = \frac{1}{9} (x,y) \in [0,3] \times [0,3],\]

    \[p(x,y) = 0 \space \text{otherwise}\]

    Find the probability that the point \((X,Y)\) is inside the unit square and interpret the result.

    53)  Consider \(X\) and \(Y\) two random variables of probability densities \(p_1(x)\) and \(p_2(x)\), respectively. The random variables \(X\) and \(Y\) are said to be independent if their joint density function is given by \(p_(x,y) = p_1(x)p_2(y)\). At a drive-thru restaurant, customers spend, on average, 3 minutes placing their orders and an additional 5 minutes paying for and picking up their meals. Assume that placing the order and paying for/picking up the meal are two independent events \(X\) and \(Y\). If the waiting times are modeled by the exponential probability densities

    \[p_1(x) = \frac{1}{3}e^{-x/3} \space x\geq 0,\]

    \[p_1(x) = 0 \space \text{otherwise}\]

    \[p_2(y) = \frac{1}{5} e^{-y/5} \space y \geq 0\]

    \[p_2(y) = 0 \space \text{otherwise}\]

    respectively, the probability that a customer will spend less than 6 minutes in the drive-thru line is given by \(P[X + Y \leq 6] = \iint_D p(x,y) dx \space dy\), where \(D = {(x,y)|x \geq 0, \space y \geq 0, \space x + y \leq 6}\). Find \(P[X + Y \leq 6]\) and interpret the result.

    Answer:
    \(P[X + Y \leq 6] = 1 + \frac{3}{2e^2} - \frac{5}{e^{6/5}} \approx 0.45\); there is a \(45\%\) chance that a customer will spend \(6\) minutes in the drive-thru line.

    54)  [T] The Reuleaux triangle consists of an equilateral triangle and three regions, each of them bounded by a side of the triangle and an arc of a circle of radius s centered at the opposite vertex of the triangle. Show that the area of the Reuleaux triangle in the following figure of side length \(s\) is \(\frac{s^2}{2}(\pi - \sqrt{3})\).

    An equilateral triangle with additional regions consisting of three arcs of a circle with radius equal to the length of the side of the triangle. These arcs connect two adjacent vertices, and the radius is taken from the opposite vertex.

    55)  [T] Show that the area of the lunes of Alhazen, the two blue lunes in the following figure, is the same as the area of the right triangle ABC. The outer boundaries of the lunes are semicircles of diameters \(AB\) and \(AC\) respectively, and the inner boundaries are formed by the circumcircle of the triangle \(ABC\).

    A right triangle with points A, B, and C. Point B has the right angle. There are two lunes drawn from A to B and from B to C with outer diameters AB and AC, respectively, and with the inner boundaries formed by the circumcircle of the triangle ABC.