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Fittoniya [83]
2 years ago
6

Select the correct answer. what makes up most hindu art? a. sculptures of the human form b. sculptures and carvings of gods c. a

rchitecture from the indus valley
SAT
1 answer:
Vedmedyk [2.9K]2 years ago
4 0

Answer:b. sculptures and carvings of gods

Explanation:

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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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A grid map marks the plot of harold’s garden in meters. The coordinates of the quadrilateral-shaped property are g(–8, 3), a(4,
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