Two question here please answer it correctly
40. Evaluate f (x2-2xy)dx +(x2y+3)dy around the boundary of the region defined by y2 = 8x and x = 2
(a) directly, (b) by using Green's theorem. Ans. 128/5
(TT.2)
41. Evaluate f (6xy - y2) dx + (3x2 --- 2xy) dy along the cycloid x = 6 - sin 6, y = 1 - cos 6.
Answers
(40) It looks like the line integral is
[tex]\displaystyle \int_C (x^2-2xy)\,\mathrm dx + (x^2y+3)\,\mathrm dy[/tex]
where C is the boundary of the region D,
[tex]D = \left\{(x,y) \mid \dfrac{y^2}8\le x\le2\text{ and }-4\le y\le4\right\}[/tex]
(a) To evaluate the line integral directly, split up C into two paths C₁ and C₂, parameterized by
• C₁ : x = t ²/8 and y = -t, where -4 ≤ t ≤ 4
(the y component is negative to make this path have a positive/counterclockwise orientation)
• C₂ : x = 2 and y = t, where -4 ≤ t ≤ 4
Then the line integral over C is the sum of the line integrals over C₁ and C₂ :
[tex]\displaystyle \int_C = \int_C \left(\left(x(t)^2-2x(t)y(t)\right)\dfrac{\mathrm dx}{\mathrm dt} + \left(x(t)^2y(t)-3\right)\dfrac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt \\\\ = \int_{-4}^4 \left(\left(\frac{t^3}4+\frac{t^4}{64}\right)\frac t4 - \left(3-\frac{t^5}{64}\right)\right)\,\mathrm dt + \int_{-4}^4 (4t+3)\,\mathrm dt \\\\ = \int_{-4}^4 \left(4t+\dfrac{t^4}{16}+\dfrac{5t^5}{256}\right)\,\mathrm dt = \boxed{\frac{128}5}[/tex]
(b) Using Green's theorem, we have
[tex]\displaystyle \int_C (x^2-2xy)\,\mathrm dx + (x^2+3)\,\mathrm dy = \iint_D \frac{\partial(x^2y+3)}{\partial x} - \frac{\partial(x^2-2xy)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_{-4}^4 \int_{y^2/8}^2 (2xy+2x)\,\mathrm dx\,\mathrm dy = \boxed{\frac{128}5}[/tex]
# # #
(41) I assume you meant to use θ in the parameterization, and not 6, so that C is parameterized by x = θ - sin(θ) and y = 1 - cos(θ). There's no range given for θ, so I'll just assume 0 ≤ θ ≤ π.
Then ... (for some reason, the math text won't render properly. I've attached the computation as an image)
(where s = sin(θ) and c = cos(θ))
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uestion 10
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Work the previous problem for the line integral f (x2+y2)dx + 3xy2 dy.
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Feel free to consult the details in my answer to 24438105 if you wish to compute the line integral directly. I don't see any specification of which method to use, so I'll do it the faster way.
By Green's theorem,
[tex]\displaystyle \int_C (x^2+y^2)\,\mathrm dx + 3xy^2\,\mathrm dy = \iint_D \frac{\partial(3xy^2)}{\partial x} - \frac{\partial(x^2+y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \iint_D (3y^2-2y)\,\mathrm dx\,\mathrm dy[/tex]
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[tex]x^2+y^2 = r^2[/tex]
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[tex]\implies \displaystyle \iint_D (3y^2-2y) \,\mathrm dx\,\mathrm dy = \int_0^{2\pi}\int_0^2 (3r^2\sin^2(t)-2r\sin(t))r\,\mathrm dr\,\mathrm dt \\\\ = \int_0^{2\pi} \int_0^2 (3r^3\sin^2(t)-2r^2\sin(t))\,\mathrm dr\,\mathrm dt \\\\ = \int_0^{2\pi} \left(12\sin^2(t)-\frac{16}3\sin(t)\right)\,\mathrm dt = \boxed{12\pi}[/tex]
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