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Mathematics LibreTexts

14.1E: Iterated Integrals and Area (Exercises)

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  • Page ID
    13708
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    Terms and Concepts

    1. When integrating \(f_x(x,y)\) with respect to \(x\), the constant of integration C is really which: \(C(x)\text{ or }C(y)\)? What does this mean?

    Answer:
    The constant of integration will be \(C(y)\) since \(y\) is considered a constant in the integration and thus, the original function \(f\) could have terms that are functions of \(y\) that were lost when taking the partial derivative with respect to the variable \(x\).

    2. Integrating an iterated integral is called _________ __________.

    Answer:
    iterated integration or multiple integration

    3. When evaluating an iterated integral of the form \(\displaystyle \int_a^b \int_{g_1(x)}^{g_2(x)}\,dy\,dx\), we integrate from _______ to ________, then from _________ to __________.

    Answer:
    We integrate from bottom to top, then from left to right.  Here, from \(y = g_1(x)\) to \(y = g_2(x)\), then from \(x = a\) to \(x = b\).
    More generally, we integrate from curve to curve and then from point to point.

    4. One understanding of an iterated integral is that \(\displaystyle \int_a^b \int_{g_1(x)}^{g_2(x)}\,dy\,dx\) gives the _______ of a plane region.

    Answer:
    area

    Problems 

    In Exercises 5-10, evaluate the integral and subsequent iterated integral.

    5. (a) \(\displaystyle \int_2^5 (6x^2+4xy-3y^2)\,dy\)
        (b) \(\displaystyle \int_{-3}^2 \int_2^5 (6x^2+4xy-3y^2)\,dy\,dx\)

    Answer:
    (a) \(\displaystyle \int_2^5 (6x^2+4xy-3y^2)\,dy\)   \(= \quad \left(6x^2y + 2xy^2 - y^3\right)\bigg|_2^5\)   \(= \quad 30x^2  + 50x - 125 - (12x^2 + 8x - 8)\)   \(=\quad 18x^2 +42x - 117\)
    (b) \(\displaystyle \int_{-3}^2 \int_2^5 (6x^2+4xy-3y^2)\,dy\,dx\)   \(=\quad \displaystyle \int_{-3}^2 \left( 18x^2 +42x - 117 \right)\, dx\)   \(=\quad \left(6x^3 + 21x^2 -117x\right)\bigg|_{-3}^2\)   \(=\quad 48 + 84 - 234 - (-162 +189 +351)\)   \(=\quad -102 - 378\)   \(=\quad 480\)

    6. (a) \(\displaystyle \int_0^\pi (2x\cos y +\sin x)\,dx\)
        (b) \(\displaystyle \int_{0}^{\pi/2} \int_0^\pi (2x\cos y +\sin x)\,dx\,dy\)

    7. (a) \(\displaystyle \int_1^x (x^2y-y+2)\,dy\)
        (b) \(\displaystyle \int_0^2 \int_1^x (x^2y-y+2)\,dy\,dx\)

    Answer:
    (a) \(\displaystyle \int_1^x (x^2y-y+2)\,dy\)   \(= \quad \left(\frac{x^2y^2}{2} - \frac{y^2}{2} + 2y\right)\bigg|_1^x\)   \(= \quad \frac{x^4}{2} - \frac{x^2}{2} + 2x - (\frac{x^2}{2} - \frac{1}{2} + 2)\)   \(=\quad \frac{x^4}{2} - x^2 + 2x - \frac{3}{2}\)
    (b) \(\displaystyle \int_0^2 \int_1^x (x^2y-y+2)\,dy\,dx\)   \(=\quad \displaystyle \int_{0}^2 \left( \frac{x^4}{2} - x^2 + 2x - \frac{3}{2} \right)\, dx\)   \(=\quad \left(\frac{x^5}{10} - \frac{x^3}{3} + x^2 - \frac{3}{2}x\right)\bigg|_{0}^2\)   \(=\quad \frac{32}{10} - \frac{8}{3} + 4 - 3 - 0\)   \(=\quad \frac{96}{30} - \frac{80}{30}  + \frac{30}{30}\)   \(=\quad \dfrac{23}{15}\)

    8. (a) \(\displaystyle \int_y^{y^2} (x-y)\,dx\)
        (b) \(\displaystyle \int_{-1}^1 \int_y^{y^2} (x-y)\,dx\,dy\)

    9. (a) \(\displaystyle \int_0^{y} (\cos x \sin y)\,dx\)
        (b) \(\displaystyle \int_0^{\pi} \int_0^{y} (\cos x \sin y)\,dx\,dy\)

    Answer:
    (a) \(\displaystyle \int_0^{y} (\cos x \sin y)\,dx\quad\) \(= \quad \left(\sin x \sin y \right)\bigg|_0^y\)   \(= \quad \sin^2 y\)
    (b) \(\displaystyle \int_0^{\pi} \int_0^{y} (\cos x \sin y)\,dx\,dy\quad\) \(=\quad \displaystyle \int_{0}^{\pi} \sin^2 y \, dy\quad\) \(=\quad \displaystyle \int_{0}^{\pi} \frac{1-\cos 2y}{2}\, dy\quad\) \(=\quad \dfrac{y - \frac{1}{2}\sin 2y}{2}\bigg|_{0}^{\pi}\quad\) \(=\quad \frac{\pi}{2} - 0\quad\) \(=\quad \dfrac{\pi}{2}\)

    10. (a) \(\displaystyle \int_0^{x} \left (\frac{1}{1+x^2}\right )\,dy\)
          (b) \(\displaystyle \int_1^2 \int_0^{x} \left (\frac{1}{1+x^2}\right )\,dy\,dx\)

    In Exercises 11-16, a graph of a planar region \(R\) is given. Give the iterated integrals, with both orders of integration \(dy\,dx\) and \(dx\,dy\), that give the area of \(R\). Evaluate one of the iterated integrals to find the area.

    11.
    13111.PNG

    Answer:
    \(\text{Area} = \displaystyle \int_1^4 \int_{-2}^1 1\,dy\,dx\quad\) \(=\quad \displaystyle \int_{-2}^1 \int_1^4  1\,dx\,dy\)
    \(\displaystyle \int_{-2}^1 \int_1^4  1\,dx\,dy\quad\) \(=\quad \displaystyle \int_{-2}^1 x\bigg|_1^4 \,dy\quad\) \(=\quad \displaystyle \int_{-2}^1 3 \,dy\quad\) \(=\quad 3y\bigg|_{-2}^1 \quad\) \(=\quad 3(3) \quad\) \(=\quad 9 \,\text{units}^2\)

    12.
    13112.PNG

    13.
    13113.PNG

    Answer:
    \( \text{Area} = \displaystyle \int_2^4 \int_{x-1}^{7-x} 1\,dy\,dx\quad\) \(=\quad \displaystyle \int_{1}^3 \int_{2}^{y+1}  1\,dx\,dy\quad + \quad \displaystyle \int_{3}^5 \int_{2}^{7-y}  1\,dx\,dy\quad\)
    \(\displaystyle \int_2^4 \int_{x-1}^{7-x} 1\,dy\,dx\quad\) \(=\quad\displaystyle \int_2^4 y\bigg|_{x-1}^{7-x}\,dx\quad\) \(=\quad\displaystyle\int_2^4 (8 - 2x)\,dx\quad\) \(=\quad(8x - x^2)\bigg|_2^4 = 32 - 16 - 16 + 4 \quad\) \(=\quad 4 \,\text{units}^2\)

    14.
    13114.PNG

    15.
    13115.PNG

    Answer:
    \( \text{Area} = \displaystyle \int_0^1 \int_{x^4}^{\sqrt{x}} 1\,dy\,dx\quad\) \(=\quad \displaystyle \int_{0}^1 \int_{y^2}^{\sqrt[4]{y}}  1\,dx\,dy\quad\)
    \( \displaystyle \int_0^1 \int_{x^4}^{\sqrt{x}} \,dy\,dx\quad\) \(=\quad\displaystyle \int_0^1 y\bigg|_{x^4}^{\sqrt{x}}\,dx\quad\) \(=\quad\displaystyle\int_0^1 (\sqrt{x} - x^4)\,dx\quad\) \(=\quad(\frac{2}{3}x^{3/2} - \frac{x^5}{5})\bigg|_0^1 \quad\) \(=\quad \frac{2}{3} - \frac{1}{5} \quad\) \(=\quad \frac{10}{15} - \frac{3}{15} \) \(=\quad \dfrac{7}{15} \,\text{units}^2\)

    16.
    13116.PNG

    In Exercises 17-22, iterated integrals are given that compute the area of a region R in the \(xy\)-plane. Sketch the region R, and give the iterated integral(s) that give the area of R with the opposite order of integration.

    17. \(\displaystyle \int_{-2}^2 \int_0^{4-x^2}\,dy\,dx\)

    Answer:
    \(\displaystyle \int_{-2}^2 \int_0^{4-x^2}\,dy\,dx\quad\)  \(=\quad \displaystyle \int_{0}^4 \int_{-\sqrt{4-y}}^{\sqrt{4-y}}\,dx\,dy\)
    Figure14-1-Ex19a.pngFigure14-1-Ex19b.png

    18. \(\displaystyle \int_{0}^1 \int_{5-5x}^{5-5x^2}\,dy\,dx\)

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

    Answer:
    \(\displaystyle \int_{-2}^2 \int_0^{2\sqrt{4-y^2}}\,dx\,dy\quad\)  \(=\quad \displaystyle \int_{0}^4 \int_{-\frac{1}{2}\sqrt{16-x^2}}^{\frac{1}{2}\sqrt{16-x^2}}\,dy\,dx\)
    Fig14-1-Ex19a.pngFig14-1-Ex19b.png

    20. \(\displaystyle \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\,dy\,dx\)

    21. \(\displaystyle \int_{0}^1 \int_{-\sqrt{y}}^{\sqrt{y}}\,dx\,dy +\int_1^4 \int_{y-2}^{\sqrt{y}}\,dx\,dy\)

    Answer:
    \(\displaystyle \int_{0}^1 \int_{-\sqrt{y}}^{\sqrt{y}}\,dx\,dy +\int_1^4 \int_{y-2}^{\sqrt{y}}\,dx\,dy \quad\) \(=\quad \displaystyle \int_{-1}^2 \int_{x^2}^{x+2}\,dy\,dx\)
    Figure14-1-Ex21a.pngFigure14-1-Ex21b.png

    22. \(\displaystyle \int_{-1}^1 \int_{(x-1)/2}^{(1-x)/2}\,dy\,dx\)