Ok, this is the second question to give my points away Might have to do third lol

SAT

2 answers:

Pepsi [2]3 years ago

7 0

Yes yes yess heheheheh

bekas [8.4K]3 years ago

4 0

<h2>ayeee tysm <3</h2><h2 /><h2>Have a great day!!</h2><h2 /><h2>~✰♥︎</h2>

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Consider the following line integral. C xy dx + x2y3 dy, c is counterclockwise around the triangle with vertices (0, 0), (1, 0),

Sedaia [141]

It looks like the integral is

$\displaystyle \int_C xy \, dx + x^2y^3 \, dy$

Let's close the loop by adding a line integral over the line segment joining (1, 4) to (0, 0). Then the closed loop is the triangular region

T = {(x, y) : 0 ≤ x ≤ 1 and 0 ≤ y ≤ 4x}

Since the integrand has no singularities over or on the boundary of T, we have by Green's theorem

$\displaystyle \int_C xy \, dx + x^2y^3 \, dy = \iint_T \left(\frac{\partial(x^2y^3)}{\partial x} - \frac{\partial(xy)}{\partial y}\right) \, dx \, dy = \int_0^1 \int_0^{4x} (2xy^3 - x) \, dy \, dx$

Compute the double integral:

$\displaystyle \int_0^1 \int_0^{4x} (2xy^3 - x) \, dy \, dx = \int_0^1 \int_0^{4x} \left(128x^5 - 4x^2\right) \, dx = \frac{60}3$

From this result, we subtract the line integral over the extra line segment we added. Parameterize this path by

C' : {(1 - t, 4 - 4t) : 0 ≤ t ≤ 1}

The line integral over C' is

$\displaystyle \int_{C'} xy \, dx + x^2y^3 \, dy = \int_0^1 (1-t)(4-4t)(-dt) + (1-t)^2(4-4t)^3(-4\,dt)$

$\displaystyle \int_{C'} xy \, dx + x^2y^3 \, dy = \int_0^1 \left(256 t^5-1280 t^4+2560 t^3-2564 t^2+1288 t-260 \right) \, dt = -44$

so that the line integral over C alone is 60/3 - (-44) = 64.

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Answer:

Explanation:

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Explanation:

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