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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Work the previous problem for the line integral f (x2+y2)dx + 3xy2 dy.
Answers
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]
Since D is a disk with radius 2 centered at the origin, consider converting to polar coordinates using x = r cos(t ) and y = r sin(t ). Then
[tex]D = \left\{(r,\theta) \mid 0\le r\le 2\text{ and }0\le t\le2\pi\right\}[/tex]
[tex]x^2+y^2 = r^2[/tex]
[tex]\mathrm dx\,\mathrm dy = r\,\mathrm dr\,\mathrm dt[/tex]
[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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Answers
Answer:
Explanation:
Generally speaking, flammable liquids will ignite (catch on fire) and burn easily at normal working temperatures. ... Fuels and many common products like solvents, thinners, cleaners, adhesives, paints, waxes and polishes may be flammable or combustible liquids.
The material which is considered flammable is: A. paint.
A chemical property can be defined as the property of a chemical compound (material) that can be observed and is measurable during a chemical reaction.
In Science, some examples of the chemical properties of a material (substance) include the following;
TarnishingRustingReactivityFlammabilityFlammability refers to the ability of a material to support combustion or burn continuously in the presence of air. An example of a material which is considered to be flammable is paint because it contains a high level of solvents.
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Answer:
The signal person is like the eyes and the equipment operator is like the hands.
Explanation:
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Answer:
no
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Answer:
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I HOPE IT WILL HELP YOU
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Answer:
C.) Disagreement about unknown facts
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Answer:
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What is the ABS system?An anti-lock braking system (ABS) is defined as an automobile safety system that allows the wheels on a motor vehicle to manage tractive contact with the road surface according to driver inputs by braking, preventing the wheels from closing up (ceasing rotation), and avoiding uncontrolled skidding.There are three basic types of anti-lock braking systems namely four-channel/four-sensor, three-channel/three-sensor, and one-channel/one-sensor.It has 4 components namely a speed sensor, valves, ECU, and Hydraulic Control Unit.To learn more about the ABS system refer to:
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Answer:
A traction control system is a secondary function of electronic stability control on production motor vehicles, designed to prevent loss of traction of the driven road wheels.
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Answer:
here
Explanation:
try not to touch the Glass on Halogen Light Bulbs, even when changing the bulb. This is because when you touch a Halogen Light Bulb, you leave behind a residue on the Light Bulb which can in time cause the bulb to heat up unevenly, and even cause the bulb to shatter as a result.
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Answer:
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Explanation:
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Plssss mark as BRAINLIST
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digital identity is the answer
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Answers
Answer:
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Ohm's law allway applies to linear constant resistances, but there is much more to an electrical system than just plain resistances.
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Answer:
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Answer:
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[tex] \frac{1}{rt} = \frac{1}{r1} + \frac{1}{r2} + \frac{1}{r3} [/tex]
If R1 = 4Ω , R2 = 6Ω and R3 = 8Ω (Ω= ohm)
Calculate: 4.1.1 Rt (series)
4.1.2 Rt (parallel)
Answers
Answer:
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[tex] \frac{1}{r} = \frac{1}{r_{1} } + \frac{1}{r _{2} } + \frac{1}{r_{3} } [/tex]
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Answer:
Recognize that there is a moral dilemma.
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Apply the resolution principles. ...
Investigate the trilemma options. ...
Make the decision.