HW 22

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Coordinate Transformation & Integration

1 : Sketch the image $S$ in the $uv$ -plane of the region $R$ in the $xy$ -plane using the given transformation. $x = 9u + 2v \quad \text {and}\quad y= 9v$

2 : Sketch the image $S$ in the $uv$ -plane of the region $R$ in the $xy$ -plane using the given transformation. $x = \dfrac {1}{2}(u+v) \quad \text {and}\quad y = \dfrac {1}{2}(u-v)$

3 : Find the Jacobian of the transformation. $x = 27uv, \quad y = \frac {4u}{v}.$ $\frac {\partial (x,y)}{\partial (u,v)} = \answer {-\frac {216 \, u}{v}}$

4 : Find the Jacobian of the transformation. $x = 2e^{-4r}\sin (5\theta ) \quad y = e^{4r}\cos (5\theta )$ $\frac {\partial (x,y)}{\partial (r,\theta )} = \answer {-40 \, \cos \left (5 \, \theta \right )^{2} + 40 \, \sin \left (5 \, \theta \right )^{2}}$

5 : Consider the region $R$ in the $xy$ -plane bounded by the ellipses $\frac {x^2}{64} + \frac {y^2}{81} = 1$ and the transformation $x = 8u$ and $y = 9v$ . Compute $\dfrac {\partial (x,y)}{\partial (u,v)}$ . $\frac {\partial (x,y)}{\partial (u,v)} = \answer {72}$

5.1 : Find the area of the ellipse using the given transformation.

Hint: The area of the unit circle is $\pi$ . $\answer {72 \, \pi }$

6 : Use the change of variables $x = u + v$ and $y = u$ to evaluate the double integral. $\iint \limits _{R} \frac {1}{4}y(x-y)\ dA,$ where $R$ is the region enclosed by the parallelogram shown below.

$\iint \limits _{R} \frac {1}{4}y(x-y)\ dA = \answer {9}$

7 : Use the change of variables $x = \dfrac {1}{2}(u + v)$ and $y = \dfrac {1}{2}(u - v)$ to set up the double integral. $\iint \limits _{R} 6 (x+y) e^{x-y}\ dA,$ where $R$ is the triangular region shown below.

$\iint \limits _{R} 6 (x+y) e^{x-y}\ dA = \int _{\answer {0}}^{\answer {2}} \int _{\answer {u - 2}}^{\answer {0}} \ \answer {3 \, u e^v}\ dv\ du$

8 : Use the given transformation $x = 7u + v$ , $y = u + 7v$ to set up the integral. $\iint \limits _{R} \frac {y}{8}\ dA,$ where $R$ is the triangular region with vertices $(0,0)$ , $(7,1)$ , and $(1,7)$ . $\iint \limits _{R} \frac {y}{8}\ dA = \int _{\answer {0}}^{\answer {1}} \int _{\answer {0}}^{\answer {1 - u}} \ \answer {6 \, u + 42 \, v}\ dv\ du$

9 : Use the given transformation $x = 4u$ , $y = 3v$ to evaluate the integral. $\iint \limits _{R} 4 \, x^{2} \ dA,$ where $R$ is the region bounded by the ellipse $\dfrac {x^2}{16} + \dfrac {y^2}{9} = 1$ . $\iint \limits _{R} 4 \, x^{2} \ dA = \answer {192 \, \pi }$

10 : Use the transformation $x = \dfrac {u}{v}$ and $y = v$ to set up the integral $\iint \limits _{R} 4xy \ dA,$ where $R$ is the region in the first quadrant bounded by the lines $y = \dfrac {x}{2}$ and $y = 2x$ , and the hyperbolas $xy = \dfrac {1}{2}$ and $xy = 2$ . $\iint \limits _{R} 4xy \ dA = \int _{\answer {\frac {1}{2}}}^{\answer {2}} \int _{\answer {\sqrt {\frac {1}{2} \, u}}}^{\answer {\sqrt {2 \, u}}} \ \answer {\frac {4 \, u}{v}}\ dv\ du$

11 : Evaluate the integral by making an appropriate change of variables. $\iint \limits _{R} 4(x+y)e^{x^2-y^2} \ dA,$ where $R$ is the rectangle enclosed by the lines $x+y = 0$ , $x + y = 4$ , $x-y = 0$ , and $x-y = 6$ . $\iint \limits _{R} 4(x+y)e^{x^2-y^2} \ dA = \answer {\frac {1}{3} \, e^{24} - \frac {25}{3}}$

12 : Evaluate the integral by making an appropriate change of variables. $\iint \limits _{R} 12\sin (16x^2 + 64y^2) \ dA,$ where $R$ is the region in the first quadrant bounded by the ellipse $16x^2 + 64y^2 = 1$ . $\iint \limits _{R} 12\sin (16x^2 + 64y^2) \ dA = \answer {-\frac {3}{32} \, \pi {\left (\cos \left (1\right ) - 1\right )}}$

13 : Use the technique of change of variable to find the volume of the solid enclosed by the ellipsoid $\frac {x^2}{9} + \frac {y^2}{4} + \frac {z^2}{25} = 1$ . Note that the volume of a unit ball is $\frac {4\pi }{3}$ . $V = \answer {40 \, \pi }$