Iterated Integrals and Area (Exercises)

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.

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.

13.

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.

15.

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.

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:

\(\displaystyle \int_{-2}^2 \int_0^{4-x^2}\,dy\,dx\quad\) \(=\quad \displaystyle \int_{0}^4 \int_{-\sqrt{4-y}}^{\sqrt{4-y}}\,dx\,dy\)

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:

\(\displaystyle \int_{-2}^2 \int_0^{2\sqrt{4-y^2}}\,dx\,dy\quad\) \(=\quad \displaystyle \int_{0}^4 \int_{-\frac{1}{2}\sqrt{16-x^2}}^{\frac{1}{2}\sqrt{16-x^2}}\,dy\,dx\)

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:

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

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