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7.2 Answer

7.2

1.

(a)

$\displaystyle \Omega = \{(x,y) : -1 \leq x \leq 1, 0 \leq y \leq 3\}$

Then we use V-simple.
$\displaystyle \iint_{\Omega} x^2 dxdy$ $\displaystyle =$ $\displaystyle \int_{x=-1}^{1}(\int_{y=0}^{3}x^2 dy)dx$  
  $\displaystyle =$ $\displaystyle \int_{-1}^{1}([x^2 y]_{y=0}^{3})dx = \int_{-1}^{1}3x^2 dx$  
  $\displaystyle =$ $\displaystyle [x^3]_{-1}^{1} = 1 - (-1) = 2$  

(b)

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

Then we use V-simple.
$\displaystyle \iint_{\Omega} e^{x+y} dxdy$ $\displaystyle =$ $\displaystyle \int_{x=0}^{1}(\int_{y=0}^{x}e^{x+y}dy)dx$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}([e^{x+y}]_{y=0}^{3})dx = \int_{0}^{1}(e^{2x} - e^{x}) dx$  
  $\displaystyle =$ $\displaystyle [\frac{e^{2x}}{2} - e^{x}]_{0}^{1} = \frac{e^{2}}{2} - e -(\frac{1}{2} - 1)$  
  $\displaystyle =$ $\displaystyle \frac{e^{2}}{2} - e + \frac{1}{2}$  

(c)

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

Then we use H-simple.
$\displaystyle \iint_{\Omega} \sqrt{xy} dxdy$ $\displaystyle =$ $\displaystyle \int_{y=0}^{1}(\int_{x=y^2}^{y}\sqrt{xy}dx)dy$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}(\sqrt{y}\int_{y^2}^{y}\sqrt{x} dx)dy = \int_{0}^{1}\sqrt{y}([\frac{2}{3}x^{\frac{3}{2}}]_{y^2}^{y} ) dy$  
  $\displaystyle =$ $\displaystyle \frac{2}{3}\int_{0}^{1}\sqrt{y}(y^{\frac{3}{2}} - y^3)dy = \frac{2}{3}\int_{0}^{1}(y^2 - y^{\frac{7}{2}})dy$  
  $\displaystyle =$ $\displaystyle \frac{2}{3}[\frac{y^3}{3} - \frac{2}{9}y^{\frac{9}{2}}]_{0}^{1} = \frac{2}{3}(\frac{1}{3} - \frac{2}{9})$  
  $\displaystyle =$ $\displaystyle \frac{2}{27}$  

(d) Find the intersection of $y^2 = 2x$ and $y^2 = 8 - 2x$. $2x = 8 - 2x$ implies $x = 2, y = \pm 2$. Now we use H-simple. Then

$\displaystyle \Omega = \{(x,y) : -2 \leq y \leq 2, \frac{y^2}{2} \leq x \leq \frac{8-y^2}{2} \}$

Thus
$\displaystyle \iint_{\Omega} (4 - y^2) dxdy$ $\displaystyle =$ $\displaystyle \int_{y=-2}^{2}(\int_{x=\frac{y^2}{2}}^{\frac{8-y^2}{2}}(4 - y^2)dx)dy$  
  $\displaystyle =$ $\displaystyle \int_{-2}^{2}([4x - y^2 x]_{\frac{y^2}{2}}^{\frac{8 - y^2}{2}} )dy$  
  $\displaystyle =$ $\displaystyle \int_{-2}^{2}(2(8-y^2) - y^2 (\frac{8 - y^2}{2}) - (2y^2 - \frac{y^4}{2}) ) dy$  
  $\displaystyle =$ $\displaystyle \int_{-2}^{2}(16 - 2y^2 - 4y^2 + \frac{y^4}{2} - 2y^2 + \frac{y^4}{2}) dy$  
  $\displaystyle =$ $\displaystyle \int_{-2}^{2}(y^4 - 8y^2 + 16)dy = 2\int_{0}^{2}(y^4 - 8y^2 + 16)dy$  
  $\displaystyle =$ $\displaystyle 2[\frac{y^5}{5} - \frac{8y^3}{3} + 16y]_{0}^{2} = 2(\frac{256}{15}) = \frac{512}{15}.$  

2.

(a) $\displaystyle{\int_{0}^{1}\int_{x^4}^{x^2}f(x,y)dydx}$. Then

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

is given by V-simple．Thus, we write $\Omega$ using H-simple.

$\displaystyle \Omega = \{(x,y) : 0 \leq y \leq 1, \sqrt{y} \leq x \leq y^{\frac{1}{4}}\} $

Thus,

$\displaystyle \int_{0}^{1}\int_{x^4}^{x^2}f(x,y) dy dx = \int_{0}^{1}\int_{\sqrt{y}}^{y^{\frac{1}{4}}}f(x,y) dx dy$

(b) $\displaystyle{\int_{0}^{1}\int_{-y}^{y}f(x,y)dxdy}$.

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

is given bu H-simple．Then we write $\Omega$ using V-simple.

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

Thus
$\displaystyle \int_{0}^{1}\int_{-y}^{y}f(x,y) dx dy$ $\displaystyle =$ $\displaystyle \int_{-1}^{0}\int_{-x}^{1}f(x,y) dy dx$  
  $\displaystyle +$ $\displaystyle \int_{0}^{1}\int_{x}^{1}f(x,y) dy dx$  

(c) $\displaystyle{\int_{1}^{4}\int_{x}^{2x}f(x,y)dydx}$.

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

is given by V-simple．Then we write $\Omega$ using H-simple.
$\displaystyle \Omega$ $\displaystyle =$ $\displaystyle \{(x,y) : 1 \leq y \leq 2, 1 \leq x \leq y\} \cup \{(x,y) : 2 \leq y \leq 4, \frac{y}{2} \leq x \leq y\}$  
  $\displaystyle \cup$ $\displaystyle \{(x,y) : 4 \leq y \leq 8, \frac{y}{2} \leq x \leq 4\}$  

Thus
$\displaystyle \int_{1}^{4}\int_{x}^{2x}f(x,y) dy dx$ $\displaystyle =$ $\displaystyle \int_{1}^{2}\int_{1}^{y}f(x,y) dx dy$  
  $\displaystyle +$ $\displaystyle \int_{2}^{4}\int_{\frac{y}{2}}^{y}f(x,y) dx dy + \int_{4}^{8}\int_{\frac{y}{2}}^{4}f(x,y) dxdy$  

3.

(a) we can't integrate $\displaystyle{\int_{0}^{1}\int_{y}^{1}e^{y/x}dxdy}$．So, we use the change of order of integration.

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

is given by H-simple．Then we erite $\Omega$ using V-simple.

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

Thus
$\displaystyle \int_{0}^{1}\int_{y}^{1}e^{y/x} dx dy$ $\displaystyle =$ $\displaystyle \int_{x=0}^{1}(\int_{y=0}^{x}e^{y/x}dy) dx$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}[xe^{y/x}]_{0}^{x} \ dx$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}(xe - x)dx = [\frac{(e-1)x^2}{2}]_{0}^{1} = \frac{e-1}{2}$  

(b) We can't integrate $\displaystyle{\int_{0}^{1}\int_{x}^{1}e^{y^{2}} dy dx}$．So, we use the change of order of integration.

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

is given by V-simple．Then we write，$\Omega$ using H-simple.

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

Thus
$\displaystyle \int_{0}^{1}\int_{x}^{1}e^{y^{2}} dy dx$ $\displaystyle =$ $\displaystyle \int_{y=0}^{1}(\int_{x=0}^{y}e^{y^{2}}dx) dy$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}[xe^{y^{2}}]_{0}^{y} \ dy$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}ye^{y^{2}} dy = [\frac{1}{2}e^{y^2}]_{0}^{1} = \frac{e-1}{2}$  

(c) We can't integrate $\displaystyle{\int_{0}^{1} dy \int_{y}^{\sqrt{y}}\frac{\sin{x}}{x}dx}$. Then we use the chage of order of integration.．

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

is given by H-simple. Then we write $\Omega$ using V-simple.

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

Thus
$\displaystyle \int_{0}^{1}dy\int_{y}^{\sqrt{y}}\frac{\sin{x}}{x} dx$ $\displaystyle =$ $\displaystyle \int_{x=0}^{1}(\int_{y=x^2}^{x}\frac{\sin{x}}{x}dy) dx$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}[y\frac{\sin{x}}{x}]_{x^2}^{x} \ dx$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}(\sin{x} - x\sin{x}) dx \ \left(\begin{array}{ll} \in... ...&= -x\cos{x} + \int{\cos{x}}dx\\ &= -x\cos{x} + \sin{x} + c \end{array}\right)$  
  $\displaystyle =$ $\displaystyle [-\cos{x} + x\cos{x} - \sin{x}]_{0}^{1}$  
  $\displaystyle =$ $\displaystyle -\cos{1} + \cos{1} - \sin{1} - (-1) = 1 - \sin{1}.$  

next up previous index
Next: Change of variables Up: MULTIPLE INTEGRATION Previous: Repeated integrals   索引