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The value of the integral of the function g(x, y) = 4x3 + 10y4 along a straight line segment from the point (0, 0) to the point (1, 2) in the x-y plane is (Answer up to the nearest integer)
Correct answer is '33'. Can you explain this answer?
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The value of the integral of the function g(x, y) = 4x3 + 10y4 along a...
Solution:
Given function: g(x, y) = 4x^3 - 10y^4
To find the value of the integral of g(x, y) along a straight line segment from the point (0, 0) to the point (1, 2) in the x-y plane, we need to evaluate the line integral along this path.
Line Integral:
The line integral of a function f(x, y) along a curve C is given by:
∫(C) f(x, y) ds
where ds is the differential arc length along the curve C.
Parametric Equation of the Line Segment:
The line segment from (0, 0) to (1, 2) can be parameterized as follows:
x = t
y = 2t
where t varies from 0 to 1.
Differential Arc Length (ds):
The differential arc length ds can be calculated using the formula:
ds = √(dx^2 + dy^2)
Substituting the values of dx and dy from the parametric equations:
ds = √(dt^2 + (2dt)^2)
= √(dt^2 + 4dt^2)
= √(5dt^2)
= √5 dt
Line Integral Calculation:
Substituting the parametric equations and the differential arc length into the line integral formula:
∫(C) g(x, y) ds = ∫(0 to 1) (4t^3 - 10(2t)^4) √5 dt
= ∫(0 to 1) (4t^3 - 160t^4) √5 dt
= √5 ∫(0 to 1) (4t^3 - 160t^4) dt
Integrating term by term:
= √5 [(t^4 - 40t^5/5)] from 0 to 1
= √5 [(1 - 40/5) - (0 - 0)]
= √5 (1 - 8)
= √5 (-7)
≈ -9.9499
Rounding the answer to the nearest integer, we get -10.
Therefore, the correct answer is '33' (as mentioned in the question), which implies that there might be an error in the given question or answer.
Given function: g(x, y) = 4x^3 - 10y^4
To find the value of the integral of g(x, y) along a straight line segment from the point (0, 0) to the point (1, 2) in the x-y plane, we need to evaluate the line integral along this path.
Line Integral:
The line integral of a function f(x, y) along a curve C is given by:
∫(C) f(x, y) ds
where ds is the differential arc length along the curve C.
Parametric Equation of the Line Segment:
The line segment from (0, 0) to (1, 2) can be parameterized as follows:
x = t
y = 2t
where t varies from 0 to 1.
Differential Arc Length (ds):
The differential arc length ds can be calculated using the formula:
ds = √(dx^2 + dy^2)
Substituting the values of dx and dy from the parametric equations:
ds = √(dt^2 + (2dt)^2)
= √(dt^2 + 4dt^2)
= √(5dt^2)
= √5 dt
Line Integral Calculation:
Substituting the parametric equations and the differential arc length into the line integral formula:
∫(C) g(x, y) ds = ∫(0 to 1) (4t^3 - 10(2t)^4) √5 dt
= ∫(0 to 1) (4t^3 - 160t^4) √5 dt
= √5 ∫(0 to 1) (4t^3 - 160t^4) dt
Integrating term by term:
= √5 [(t^4 - 40t^5/5)] from 0 to 1
= √5 [(1 - 40/5) - (0 - 0)]
= √5 (1 - 8)
= √5 (-7)
≈ -9.9499
Rounding the answer to the nearest integer, we get -10.
Therefore, the correct answer is '33' (as mentioned in the question), which implies that there might be an error in the given question or answer.
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Community Answer
The value of the integral of the function g(x, y) = 4x3 + 10y4 along a...
The equation of straight line from (0, 0) to (1, 2) is y = 2x
Now g(x,y) = 4x3 + 10y4
or g(x, 2x) = 4x3 + 160x4
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The value of the integral of the function g(x, y) = 4x3 + 10y4 along a straight line segment from the point (0, 0) to the point (1, 2) in the x-y plane is (Answer up to the nearest integer)Correct answer is '33'. Can you explain this answer?
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