Repeated Integrals

Figure 5.5: Vertically Simple Region

Figure 5.6: Horizontal simple

The closed region $\Omega$ bounded by $a \leq x \leq b, \phi_{1}(x) \leq y \leq \phi_{2}(x)$ is called Vertically simple region, The closed region $\Omega$ bounded by $c \leq y \leq d, \psi_{1}(y) \leq x \leq \psi_{2}(y)$ is called Horizontally simple region.

Vertically Simple

If a vertical line does not cross the same curve more than once.

Horizontal Simple

If a horizontal line does not cross the same curve more than once.

Theorem 5..3   1. If $f(x,y)$ is continuous on the closed region $\Omega(\ref{fig:V-simple})$ bounded by $a \leq x \leq b, \phi_{1}(x) \leq y \leq \phi_{2}(x)$, then

$$\iint_{\Omega}f(x,y)dxdy = \int_{a}^{b}\{\int_{\phi_{1}(x)}^{\phi_{2}(x)}f(x,y)dy\}dx$$

2. If $f(x,y)$ is continuous on the closed region $\Omega(\ref{fig:V-simple})$ bounded by $c \leq y \leq d, \psi_{1}(y) \leq x \leq \psi_{2}(y)$, then

$$\iint_{\Omega}f(x,y)dxdy = \int_{c}^{d}\{\int_{\psi_{1}(y)}^{\psi_{2}(y)}f(x,y)dx\}dy$$

Evaluation

To evaluate a double integral, use either vertically simple region or horizontally simple region.

Example 5..2   Given the region $\Omega$ as in the figure 5.7, evaluate the following double integral

$$\iint_{\Omega}(x^2 - y)dxdy$$

When a line is drawed vertically, it will not intersect the same curve more thatn once. Thus the region is vertically simple region.

Figure 5.7: $-x^2 \leq y \leq x^2$

SOLUTION When a horizontal line is drawed to the region $\Omega$, it intersects the curve more than once. But when a vertical line is drawed to the region $\Omega$, it does not intersect more than once. Thus the region is vertically simple region. Now fix $x$. Then the region is in between the curve $y = -x^2$ and the curve $y = x^2$. Thus we have $-x^2 \leq y \leq x^2$. Next free $x$ to get $-1 \leq x \leq 1$. Thus $\Omega$ is expressed as follows.

Figure 5.8: Example5-2

$$\Omega = \{(x,y) : -1 \leq x \leq 1, -x^2 \leq y \leq x^2 \}$$

Therefore,

$$\iint_{\Omega}(x^2 - y)dxdy = \int_{-1}^{1}(\int_{-x^2}^{x^2}(x^2 - y)dy)dx$$

$$= \int_{-1}^{1}\left[(x^2 y - \frac{1}{2}y^2) \right ]_{-x^2}^{x^2} dx$$

$$= \int_{-1}^{1}[(x^4 - \frac{1}{2}x^4) - (-x^4 - \frac{1}{2}x^4)]dx$$

$$= \int_{-1}^{1}2x^4dx = \left[\frac{2}{5}x^5 \right ]_{-1}^{1} = \frac{4}{5} \ensuremath{ \blacksquare}$$

Exercise 5..2   Give the region $\Omega$ as in the figure 5.9, evaluate the following double integral.

$$\iint_{\Omega}(x^{1/2} - y^2)dxdy$$

Figure 5.9: $x^2 \leq y \leq x^{1/4}$

Figure 5.10: Exercise5-2

In the above figure, summing the small rectangles to the $y$-axis direction. Then $x^2 \leq y \leq x^{1/4}$. Now to fill the region using these vertically long rectangles, we need to sum $0 \leq x \leq 1$.

SOLUTION This region is both vertically simple region and horizontally simple region. We first evaluate the integral by using vertically simple region.

$\Omega$ can be expressed by the following.

$$\Omega = \{(x,y) : 0 \leq x \leq 1, x^2 \leq y \leq x^{1/4} \}$$

Thus

$$\iint_{\Omega}(x^{1/2} - y^2)dxdy = \int_{0}^{1}\int_{x^2}^{x^{1/4}}(x^{1/2} - y^2)dydx$$

$$= \int_{0}^{1}\left[x^{1/2}y - \frac{1}{3}y^3\right ]_{x^2}^{x^{1/4}}dx$$

$$= \int_{0}^{1}(\frac{2}{3}x^{3/4} - x^{5/2} + \frac{1}{3}x^{6})dx$$

$$= \left[\frac{8}{21}x^{7/4} - \frac{2}{7}x^{7/2} + \frac{1}{21}x^7\right ]_{0}^{1}$$

$$= \frac{8}{21} - \frac{2}{7} + \frac{1}{21} = \frac{1}{7}$$

This time, we evaluate the integral by using horizontally simple region. $\Omega$ can be expressed by the following.

Figure 5.11: Example5-2

By horizontally simple region, $y^{4} \leq x \leq y^{1/2}$. Then to fill the region $\Omega$, we get $0 \leq y \leq 1$.

$$\Omega = \{(x,y) : 0 \leq y \leq 1, y^4 \leq x \leq y^{1/2} \}$$

Then

$$\iint_{\Omega}(x^{1/2} - y^2)dxdy = \int_{0}^{1}\int_{y^4}^{y^{1/2}}(x^{1/2} - y^2)dxdy$$

$$= \int_{0}^{1}\left[\frac{2}{3}x^{3/2} - y^2 x\right ]_{y^4}^{y^{1/2}}dy$$

$$= \int_{0}^{1}(\frac{2}{3}y^{3/4} - y^{5/2} + \frac{1}{3}y^{6})dy$$

$$= \left[\frac{8}{21}y^{7/4} - \frac{2}{7}y^{7/2} + \frac{1}{21}y^7\right ]_{0}^{1}$$

$$= \frac{8}{21} - \frac{2}{7} + \frac{1}{21} = \frac{1}{7}$$

We obtain the same result as expected $\blacksquare$

Interchage from vertically simple region to horizontally simple region or vice versa. Then corresponding integral change the order of integration.

Example 5..3   Evaluate the following double integral.

$$\int_{0}^{1}\int_{y}^{1}e^{y/x}dxdy$$

SOLUTION We can not evaluate the integral $${\int_{y}^{1}e^{y/x} dx}$$. Then by the change the order of integration, the region $\Omega$ is given by

$$\Omega = \{(x,y) : 0 \leq y \leq 1, y \leq x \leq 1\}$$

Thus it is given by the horizontally simple region. Now express $\Omega$ by the vertically simple region.

$$\Omega = \{(x,y) : 0 \leq x \leq 1, 0 \leq y \leq x\}$$

Then

Figure 5.12: Vertically simple

$\int e^{y/x}dy = xe^{y/x}$

$$\int_{0}^{1}\int_{y}^{1}e^{y/x}dxdy = \int_{x=0}^{1}(\int_{y=0}^{x}e^{y/x}dy) dx$$

$$= \int_{0}^{1}[xe^{y/x}]_{0}^{x} dx$$

$$= \int_{0}^{1}(xe - x)dx = \left[\frac{(e-1)x^2}{2}\right]_{0}^{1} = \frac{e-1}{2}\ensuremath{ \blacksquare}$$

Exercise 5..3   Evaluate the following double integral.

$$\int_{0}^{1}\int_{x}^{1}e^{y^2}dydx$$

$\int e^{y^2}dy$ is known for non-integrable function. Express $\int_{0}^{1}\int_{x}^{1}e^{y^{2}} dy dx$ by $\int_{0}^{1}\big(\int_{x}^{1}e^{y^{2}} dy\big) dx$. Then the range of the integration of $x$ and the range of integration of $y$ become clear.

Figure 5.13: Exercise5-3

SOLUTION Note that $\int e^{y^2}dy$ is known for non-integrable. Thus it is impossible to integrate $${\int_{0}^{1}\int_{x}^{1}e^{y^{2}} dy dx}$$ in this order. Thus, interchange the order of integration. Since

$$\Omega = \{(x,y) : 0 \leq x \leq 1, x \leq y \leq 1\}$$

is given by the vertically simple region. Then express $\Omega$ by the horizontally simple region.

$$\Omega = \{(x,y) : 0 \leq y \leq 1, 0 \leq x \leq y\}$$

Thus,

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

$$= \int_{0}^{1}[xe^{y^{2}}]_{0}^{y} dy$$

$$= \int_{0}^{1}ye^{y^{2}} dy = \left[\frac{1}{2}e^{y^2}\right]_{0}^{1} = \frac{e-1}{2}\ensuremath{ \blacksquare}$$

Exercise A

1.

Evaluate the following double integrals．

(a) $${\iint_{\Omega}x dxdy, \Omega: -1 \leq x \leq 1, 0 \leq y \leq 3}$$

(b) $${\iint_{\Omega}(2x + 3y) dxdy, \Omega: 0 \leq x \leq 1, 0 \leq y \leq x}$$

(c) $${\iint_{\Omega}(1 + x + xy) dxdy, \Omega: 0 \leq y \leq 1, y^2 \leq x \leq y}$$

(d) $${\iint_{\Omega}\sin(x+y) dxdy, \Omega: 0 \leq x \leq \frac{\pi}{2}, 0 \leq y \leq \frac{\pi}{2}}$$

(e) $${\iint_{\Omega}x^{3} y dxdy, \Omega: 0 \leq x \leq 1, 0 \leq y \leq x}$$

2.

Interchange the order of integration．

(a) $${\int_{0}^{1}\int_{x^{2}}^{x}f(x,y)dydx}$$ (b) $${\int_{0}^{1}\int_{0}^{y^{2}}f(x,y)dxdy}$$ (c) $${\int_{0}^{2}\int_{\frac{x}{2}}^{3-x}f(x,y)dydx}$$

3.

Answer the following questions．

(a) Find the volume of solid bounded by the following surface under the surface $z = x+y$ and above the triqngle $(0,0), (0,1), (1,0)$

(b) Find the volume of solid bounded by the following surface under the surface $z = 2x+3y$ and qbove the square $(0,0), (0,1), (1,0), (1,1)$.

(c) Find the volume of the solid bounded above by the surface $z = x^{2} + y^{2}$ and below by the plane $x^{2} + y^{2} \leq 1$

Exercise B

1.

Evaluate the following double integrals．

(a) $${\iint_{\Omega}x^2 dxdy, \Omega: -1 \leq x \leq 1, 0 \leq y \leq 3}$$

(b) $${\iint_{\Omega}e^{x+y} dxdy, \Omega: 0 \leq x \leq 1, 0 \leq y \leq x}$$

(c) $${\iint_{\Omega}\sqrt{xy} dxdy, \Omega: 0 \leq y \leq 1, y^2 \leq x \leq y}$$

(d) $${\iint_{\Omega}(4 - y^2) dxdy, \Omega}$$ is bounded by $y^2 = 2x$ and $y^2 = 8 - 2x$.

2.

Interchange the order of integrals.

(a) $${\int_{0}^{1}\int_{x^4}^{x^2}f(x,y)dydx}$$ (b) $${\int_{0}^{1}\int_{-y}^{y}f(x,y)dxdy}$$ (c) $${\int_{1}^{4}\int_{x}^{2x}f(x,y)dydx}$$

3.

Evaluate the following double integrals．

(a) $${\int_{0}^{1}\int_{y}^{1}e^{y/x}dxdy}$$ (b) $${\int_{0}^{1}\int_{x}^{1}e^{y^2}dydx}$$ (c) $${\int_{0}^{1} dy \int_{y}^{\sqrt{y}}\frac{\sin{x}}{x}dx}$$