15.7E: Mazoezi ya Sehemu ya 15.7

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Katika mazoezi ya 1 - 6, kazi$T : S \rightarrow R, \space T (u,v) = (x,y)$ kwenye kanda$S = \big\{(u,v) \,|\, 0 \leq u \leq 1, \space 0 \leq v \leq 1\big\}$ iliyofungwa na mraba wa kitengo hutolewa, wapi$R \in R^2$ picha ya$S$ chini$T$.

kuhalalisha kwamba kazi$T$ ni$C^1$ mabadiliko.

pata picha za vipeo vya mraba wa kitengo$S$ kupitia kazi$T$.

c Tambua picha ya mraba$R$ wa kitengo$S$ na uifanye grafu.

1. $x = 2u, \space y = 3v$

2. $x = \frac{u}{2}, \space y = \frac{v}{3}$

Jibu

a.$T(u,v) = (g(u,v), \space h(u,v), \space x = g(u,v) = \frac{u}{2}$ na$y = h(u,v) = \frac{v}{3}$. Kazi$g$ na$h$ zinaendelea na kutofautishwa, na derivatives ya sehemu$g_u (u,v) = \frac{1}{2}, \space g_v (u,v) = 0, \space h_u (u,v) = 0$ na$h_v (u,v) = \frac{1}{3}$ zinaendelea$S$;

b.$T(0,0) = (0,0), \space T(1,0) = \left(\frac{1}{2},0\right), \space T(0,1) = \left(0,\frac{1}{3}\right)$, na$T(1,1) = \left(\frac{1}{2}, \frac{1}{3} \right)$;

c.$R$ ni mstatili wa vipeo$(0,0), \space \left(0,\frac{1}{3}\right), \space \left(\frac{1}{2}, \frac{1}{3} \right)$, na$\left(0,\frac{1}{3}\right)$ katika$xy$ -ndege; takwimu inayofuata.

$Mstatili na kona moja kwa asili, urefu wa usawa 0.5, na urefu wa wima 0.34.$

3. $x = u - v, \space y = u + v$

4. $x = 2u - v, \space y = u + 2v$

Answer

a. $T(u,v) = (g(u,v), \space h(u,v), \space x = g(u,v) = 2u - v$ and $y = h(u,v) = u + 2v$. The functions $g$ and $h$ are continuous and differentiable, and the partial derivatives $g_u (u,v) = 2, \space g_v (u,v) = -1, \space h_u (u,v) = 1$ and $h_v (u,v) = 2$ are continuous on $S$;

b. $T(0,0) = (0,0), \space T(1,0) = (2,1), \space T(0,1) = (-1,2)$, and $T(1,1) = (1,3)$;

c. $R$ is the parallelogram of vertices $(0,0), \space (2,1) \space (1,3)$, and $(-1,2)$ in the $xy$-plane; the following figure.

$A square of side length square root of 5 with one corner at the origin and another at (2, 1).$

5. $x = u^2, \space y = v^2$

6. $x = u^3, \space y = v^3$

Jibu

a.$T(u,v) = (g(u,v), \space h(u,v), \space x = g(u,v) = u^3$ na$y = h(u,v) = v^3$. Kazi$g$ na$h$ zinaendelea na kutofautishwa, na derivatives ya sehemu$g_u (u,v) = 3u^2, \space g_v (u,v) = 0, \space h_u (u,v) = 0$ na$h_v (u,v) = 3v^2$ zinaendelea$S$;

b.$T(0,0) = (0,0), \space T(1,0) = (1,0), \space T(0,1) = (0,1)$, na$T(1,1) = (1,1)$;

c.$R$ ni kitengo mraba katika$xy$ -ndege kuona takwimu katika jibu kwa zoezi uliopita.

Katika mazoezi 7 - 12, onyesha kama mabadiliko$T : S \rightarrow R$ ni moja kwa moja au la.

7. $x = u^2, \space y = v^2$,$S$ wapi mstatili wa vipeo$(-1,0), \space (1,0), \space (1,1)$, na$(-1,1)$.

8. $x = u^4, \space y = u^2 + v$,$S$ wapi pembetatu ya vipeo$(-2,0), \space (2,0)$, na$(0,2)$.

Jibu: $T$si moja kwa moja: pointi mbili za$S$ kuwa na picha sawa. Hakika,$T(-2,0) = T(2,0) = (16,4)$.

9. $x = 2u, \space y = 3v$,$S$ wapi mraba wa vipeo$(-1,1), \space (-1,-1), \space (1,-1)$, na$(1,1)$.

10. $T(u, v) = (2u - v, u),$$S$wapi pembetatu yenye vipeo$(-1,1), \, (-1,-1)$, na$(1,-1)$.

Jibu: $T$ni moja kwa moja: Sisi wanasema na utata. $T(u_1,v_1) = T(u_2,v_2)$ina maana$2u_1 - v_1 = 2u_2 - v_2$ na$u_1 = u_2$. Hivyo,$u_1 = u+2$ na$v_1 = v_2$.

11. $x = u + v + w, \space y = u + v, \space z = w$, wapi$S = R = R^3$.

12. $x = u^2 + v + w, \space y = u^2 + v, \space z = w$, wapi$S = R = R^3$.

Jibu: $T$si moja kwa moja:$T(1,v,w) = (-1,v,w)$

Katika mazoezi 13 - 18, mabadiliko$T : R \rightarrow S$ ni moja kwa moja. Kupata kuhusiana mabadiliko yao inverse$T^{-1} : R \rightarrow S$.

13. $x = 4u, \space y = 5v$, wapi$S = R = R^2$.

14. $x = u + 2v, \space y = -u + v$, wapi$S = R = R^2$.

Jibu: $u = \frac{x-2y}{3}, \space v= \frac{x+y}{3}$

15. $x = e^{2u+v}, \space y = e^{u-v}$, wapi$S = R^2$ na$R = \big\{(x,y) \,|\, x > 0, \space y > 0\big\}$

16. $x = \ln u, \space y = \ln(uv)$, wapi$S = \big\{(u,v) \,|\, u > 0, \space v > 0\big\}$ na$R = R^2$.

Jibu: $u = e^x, \space v = e^{-x+y}$

17. $x = u + v + w, \space y = 3v, \space z = 2w$, wapi$S = R = R^3$.

18. $x = u + v, \space y = v + w, \space z = u + w$, wapi$S = R = R^3$.

Jibu: $u = \frac{x-y+z}{2}, \space v = \frac{x+y-z}{2}, \space w = \frac{-x+y+z}{2}$

Katika mazoezi 19 - 22, mabadiliko$T : S \rightarrow R, \space T (u,v) = (x,y)$ na kanda$R \subset R^2$ hutolewa. Pata kanda$S \subset R^2$.

19. $x = au, \space y = bv, \space R = \big\{(x,y) \,|\, x^2 + y^2 \leq a^2 b^2\big\}$wapi$a,b > 0$

20. $x = au, \space y = bc, \space R = \big\{(x,y) \,|\, \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1\big\}$, wapi$a,b > 0$

Jibu: $S = \big\{(u,v) \,|\, u^2 + v^2 \leq 1\big\}$

21. $x = \frac{u}{a}, \space y = \frac{v}{b}, \space z = \frac{w}{c}, \space R = \big\{(x,y)\,|\,x^2 + y^2 + z^2 \leq 1\big\}$, wapi$a,b,c > 0$

22. $x = au, \space y = bv, \space z = cw, \space R = \big\{(x,y)\,|\,\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} \leq 1, \space z > 0\big\}$, wapi$a,b,c > 0$

Jibu: $R = \big\{(u,v,w)\,|\,u^2 - v^2 - w^2 \leq 1, \space w > 0\big\}$

Katika mazoezi 23 - 32, tafuta Jacobian$J$ ya mabadiliko.

23. $x = u + 2v, \space y = -u + v$

24. $x = \frac{u^3}{2}, \space y = \frac{v}{u^2}$

Jibu: $\frac{3}{2}$

25. $x = e^{2u-v}, \space y = e^{u+v}$

26. $x = ue^v, \space y = e^{-v}$

Jibu: $-1$

27. $x = u \space \cos (e^v), \space y = u \space \sin(e^v)$

28. $x = v \space \sin (u^2), \space y = v \space \cos(u^2)$

Jibu: $2uv$

29. $x = u \space \cosh v, \space y = u \space \sinh v, \space z = w$

30. $x = v \space \cosh \left(\frac{1}{u}\right), \space y = v \space \sinh \left(\frac{1}{u}\right), \space z = u + w^2$

Jibu: $\frac{v}{u^2}$

31. $x = u + v, \space y = v + w, \space z = u$

32. $x = u - v, \space y = u + v, \space z = u + v + w$

Jibu: $2$

33. Mkoa wa triangular$R$ na vertices$(0,0), \space (1,1)$, na$(1,2)$ inavyoonekana katika takwimu ifuatayo.

$Pembetatu yenye pembe za asili, (1, 1), na (1, 2).$

a Kupata mabadiliko$T : S \rightarrow R, \space T(u,v) = (x,y) = (au + bv + dv)$, where $a,b,c$, and $d$ are real numbers with $ad - bc \neq 0$ such that $T^{-1} (0,0) = (0,0), \space T^{-1} (1,1) = (1,0)$, and $T^{-1}(1,2) = (0,1)$.

b. Use the transformation $T$ to find the area $A(R)$ of the region $R$.

34. The triangular region $R$ with the vertices $(0,0), \space (2,0)$, and $(1,3)$ is shown in the following figure.

$A triangle with corners at the origin, (2, 0), and (1, 3).$

a. kupata mabadiliko$T : S \rightarrow R, \space T(u,v) = (x,y) = (au + bv + dv)$, ambapo$a,b,c$, na$d$ ni namba halisi na$ad - bc \neq 0$ vile kwamba$T^{-1} (0,0) = (0,0), \space T^{-1} (2,0) = (1,0)$, na$T^{-1}(1,3) = (0,1)$.

b Tumia mabadiliko$T$ ili kupata eneo$A(R)$ la kanda$R$.

Jibu: a.$T (u,v) = (2u + v, \space 3v)$
b. eneo la$R$ ni$\displaystyle A(R) = \int_0^3 \int_{y/3}^{(6-y)/3} \, dx \, dy = \int_0^1 \int_0^{1-u} \left|\frac{\partial (x,y)}{\partial (u,v)}\right| \, dv \space du = \int_0^1 \int_0^{1-u} 6 \, dv \, du = 3.$

Katika mazoezi 35 - 36, tumia mabadiliko$u = y - x, \space v = y$, kutathmini integrals juu ya parallelogram$R$ ya vertices$(0,0), \space (1,0), \space (2,1)$, na$(1,1)$ inavyoonekana katika takwimu zifuatazo.

$Rhombus yenye pembe katika asili, (1, 0), (1, 1), na (2, 1).$

35. $\displaystyle \iint_R (y - x) \, dA$

36. $\displaystyle \iint_R (y^2 - xy) \, dA$

Answer: $-\frac{1}{4}$

In exercises 37 - 38, use the transformation $y = x = u, \space x + y = v$ to evaluate the integrals on the square $R$ determined by the lines $y = x, \space y = -x + 2, \space y = x + 2$, and $y = -x$ shown in the following figure.

$A square with side lengths square root of 2 rotated 45 degrees with one corner at the origin and another at (1, 1).$

37. $\displaystyle \iint_R e^{x+y} \, dA$

38. $\displaystyle \iint_R \sin (x - y) \, dA$

Jibu: $-1 + \cos 2$

Katika mazoezi 39 - 40, tumia mabadiliko$x = u, \space 5y = v$ ili kutathmini viungo kwenye kanda$R$ iliyofungwa na duaradufu$x^2 + 25y^2 = 1$ iliyoonyeshwa kwenye takwimu ifuatayo.

$duaradufu na kituo katika asili, mhimili kuu 2, na ndogo 0.4.$

39. $\displaystyle \iint_R \sqrt{x^2 + 25y^2} \, dA$

40. $\displaystyle \iint_R (x^2 + 25y^2)^2 \, dA$

Answer: $\frac{\pi}{15}$

In exercises 41 - 42, use the transformation $u = x + y, \space v = x - y$ to evaluate the integrals on the trapezoidal region $R$ determined by the points $(1,0), \space (2,0), \space (0,2)$, and $(0,1)$ shown in the following figure.

$A trapezoid with corners at (1, 0), (0, 1), (0, 2), and (2, 0).$

41. $\displaystyle \iint_R (x^2 - 2xy + y^2) \space e^{x+y} \, dA$

42. $\displaystyle \iint_R (x^3 + 3x^2y + 3xy^2 + y^3) \, dA$

Jibu: $\frac{31}{5}$

43. Sekta ya annulus ya mviringo$R$ imefungwa$4x^2 + 4y^2 = 1$ na miduara na$9x^2 + 9y^2 = 64$, mstari$x = y \sqrt{3}$, na$y$ -axis inavyoonekana katika takwimu ifuatayo. Pata mabadiliko$T$ kutoka mkoa wa mstatili$S$ katika$r\theta$ -ndege hadi kanda$R$ katika$xy$ -ndege. Grafu$S$.

$Katika roboduara ya kwanza, sehemu ya annulus iliyoelezwa na radius ya ndani ya 0.5, radius ya nje kidogo zaidi ya 2.5, na katikati ya asili. Kuna mstari unaogawanya annulasi hii inayotokana na takriban angle ya shahada 30. Sehemu inayohusiana na digrii 60 ni kivuli.$

44. Mango$R$ bounded by the circular cylinder $x^2 + y^2 = 9$ and the planes $z = 0, \space z = 1, \space x = 0$, and $y = 0$ is shown in the following figure. Find a transformation $T$ from a cylindrical box $S$ in $r\theta z$-space to the solid $R$ in $xyz$-space.

$A quarter of a cylinder with height 1 and radius 3. The center axis is the z axis.$

Jibu: $T (r,\theta,z) = (r \space \cos \theta, \space r \space \sin \theta, \space z); \space S = [0,3] \times [0,\frac{\pi}{2}] \times [0,1]$katika$r\theta z$ -nafasi

45. Onyesha kwamba\[\iint_R f \left(\sqrt{\frac{x^2}{3} + \frac{y^2}{3}}\right) dA = 2 \pi \sqrt{15} \int_0^1 f (\rho) \rho \space d\rho, \nonumber \] ambapo$f$ ni kazi ya kuendelea juu$[0,1]$ na$R$ ni kanda imepakana na duaradufu$5x^2 + 3y^2 = 15$.

46. Onyesha kwamba\[\iiint_R f \left(\sqrt{16x^2 + 4y^2 + z^2}\right) dV = \frac{\pi}{2} \int_0^1 f (\rho) \rho^2 d\rho, \nonumber \] ambapo$f$ ni kazi ya kuendelea juu$[0,1]$ na$R$ ni kanda imepakana na ellipsoid$16x^2 + 4y^2 + z^2 = 1$.

47. [T] Kupata eneo la mkoa imepakana na curves$xy = 1, \space xy = 3, \space y = 2x$, na$y = 3x$ kwa kutumia mabadiliko$u = xy$ na$v = \frac{y}{x}$. Tumia mfumo wa algebra ya kompyuta (CAS) ili kuchora mipaka ya mipaka ya kanda$R$.

48. [T] Kupata eneo la mkoa imepakana na curves$x^2y = 2, \space x^2y = 3, \space y = x$, na$y = 2x$ kwa kutumia mabadiliko$u = x^2y$ na$v = \frac{y}{x}$. Tumia CAS kwa graph curves mipaka ya kanda$R$.

Jibu

eneo la$R$ ni$10 - 4\sqrt{6}$; curves mipaka ya$R$ ni graphed katika takwimu zifuatazo.

$Mstari minne hutolewa, yaani, y = 3, y = 2, y = 3/ (x squared), na y = 2/ (x squared). Mstari y = 3 na y = 2 ni sawa na kila mmoja. Mstari y = 3/ (x squared) na y = 2/ (x squared) ni curves kwamba kukimbia kiasi fulani sambamba na kila mmoja.$

49. Tathmini muhimu mara tatu\[\int_0^1 \int_1^2 \int_z^{z+1} (y + 1) \space dx \space dy \space dz \nonumber \] by using the transformation $u = x - z, \space v = 3y$, and $w = \frac{z}{2}$.

50. Evaluate the triple integral \[\int_0^2 \int_4^6 \int_{3z}^{3z+2} (5 - 4y) \space dx \space dy \space dz \nonumber \] by using the transformation $u = x - 3z, \space v = 4y$, and $w = z$.

Answer: $8$

51. A transformation $T : R^2 \rightarrow R^2, \space T (u,v) = (x,y)$ of the form $x = au + bv, \space y = cu + dv$, where $a,b,c$, and $d$ are real numbers, is called linear. Show that a linear transformation for which $ad - bc \neq 0$ maps parallelograms to parallelograms.

52. A transformation $T_{\theta} : R^2 \rightarrow R^2, \space T_{\theta} (u,v) = (x,y)$ of the form $x = u \space \cos \theta - v \space \sin \theta, \space y = u \space \sin \theta + v \space \cos \theta$, is called a rotation angle $\theta$. Show that the inverse transformation of $T_{\theta}$ satisfies $T_{\theta}^{-1} = T_{-\theta}$ where $T_{-\theta}$ is the rotation of angle $-\theta$.

53. [T] Find the region $S$ in the $uv$-plane whose image through a rotation of angle $\frac{\pi}{4}$ is the region $R$ enclosed by the ellipse $x^2 + 4y^2 = 1$. Use a CAS to answer the following questions.

a. Graph the region $S$.

b. Evaluate the integral $\displaystyle \iint_S e^{-2uv} \, du \, dv.$ Round your answer to two decimal places.

54. [T] The transformations $T_i : \mathbb{R}^2 \rightarrow \mathbb{R}^2, \space i = 1, . . . , 4,$ defined by $T_1(u,v) = (u,-v), \space T_2 (u,v) = (-u,v), \space T_3 (u,v) = (-u, -v)$, and $T_4 (u,v) = (v,u)$ are called reflections about the $x$-axis, $y$-axis origin, and the line $y = x$, respectively.

a. Find the image of the region $S = \big\{(u,v)\,|\,u^2 + v^2 - 2u - 4v + 1 \leq 0\big\}$ in the $xy$-plane through the transformation $T_1 \circ T_2 \circ T_3 \circ T_4$.

b. Use a CAS to graph $R$.

c. Evaluate the integral $\displaystyle \iint_S \sin (u^2) \, du \, dv$ by using a CAS. Round your answer to two decimal places.

Answer

a. $R = \big\{(x,y)\,|\,y^2 + x^2 - 2y - 4x + 1 \leq 0\big\}$;
b. $R$ is graphed in the following figure;

$A circle with radius 2 and center (2, 1).$

c.$3.16$

55. [T] mabadiliko$T_{k,1,1} : \mathbb{R}^3 \rightarrow \mathbb{R}^3, \space T_{k,1,1}(u,v,w) = (x,y,z)$ ya fomu$x = ku, \space y = v, \space z = w$, ambapo$k \neq 1$ ni chanya idadi halisi, inaitwa kunyoosha kama$k > 1$ na compression kama$0 < k < 1$ katika$x$ -direction. Matumizi CAS kutathmini muhimu$\displaystyle \iiint_S e^{-(4x^2+9y^2+25z^2)} \, dx \, dy \, dz$ juu ya imara$S = \big\{(x,y,z) \,|\, 4x^2 + 9y^2 + 25z^2 \leq 1\big\}$ kwa kuzingatia compression$T_{2,3,5}(u,v,w) = (x,y,z)$ inavyoelezwa na$x = \frac{u}{2}, \space y = \frac{v}{3}$, na$z = \frac{w}{5}$. Pindisha jibu lako kwenye maeneo manne ya decimal.

56. [T] mabadiliko$T_{a,0} : \mathbb{R}^2 \rightarrow \mathbb{R}^2, \space T_{a,0} (u,v) = (u + av, v)$, ambapo$a \neq 0$ ni idadi halisi, inaitwa shear katika$x$ -direction. Mabadiliko$T_{b,0} : R^2 \rightarrow R^2, \space T_{o,b}(u,v) = (u,bu + v)$, wapi$b \neq 0$ namba halisi, inaitwa shear katika$y$ -mwelekeo.

a. kupata mabadiliko$T_{0,2} \circ T_{3,0}$.

pata picha ya mkoa$R$ wa$S$ trapezoidal uliowekwa na$u = 0, \space v = 0, \space v = 1$, na$v = 2 - u$ kupitia mabadiliko$T_{0,2} \circ T_{3,0}$.

c Matumizi CAS kwa graph picha$R$ katika$xy$ -ndege.

pata eneo la kanda$R$ kwa kutumia eneo la kanda$S$.

Jibu

a$T_{0,2} \circ T_{3,0}(u,v) = (u + 3v, 2u + 7v)$;.

b. picha$S$ ni quadrilateral ya vertices$(0,0), \space (3,7), \space (2,4)$, na$(4,9)$;

c.$S$ imewekwa katika takwimu ifuatayo;

$Takwimu nne na pointi asili, (2, 4), (4, 9), na (3, 7).$

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