Question

Verify that the given vector field h is a gradient. Then calculate the line integral of h over the indicated curve C by finding f such that the gradient f = h and evaluating f at the end points of C.

h(x,y) = xy^2 i + yx^2 j

r(u) = u i + 2u^2 j

0 ? u ? 1

Please show all the steps so that I will know where i am making a mistake. I have attempted this question 4 times and i do not know where am going wrong. The correct answer is 2. I keep getting 4.

Answer 1

Answer:

$f(x,y) = \frac{x^2 y^2}{2}$

The value of the line integral is 2

Step-by-step explanation:

$h(x,y) = xy^2 i + yx^2 j$

Note that if you derivate the first part over the variable y and the second part over the variable x, then in both cases you obtain 2xy, therefore there must be a function f whose gradient is h, because the cross derivates are equal.

In order to find such f, you can calculate a primitive of both expressions, the first one over the variable x and the second one over the variable y.

A general primitive of xy² i (over x) is

$f_1(x,y) = \frac{x^2y^2}{2} + a(y)$

With a(y) a function that depends only on y. A general primitive of yx² j (over y) is

$f_2(x,y) = \frac{x^2y^2}{2} + b(x)$

With b(x) only depending on x

The function f(x,y) whose gradient is h is obtained by equaling the expressions of f₁ and f₂. f₁ and f₂ are equal when a(x) = b(x) = 0, therefore

$f(x,y) = \frac{x^2y^2}{2}$

note that

• fx(x,y) = xy²
• fy(x,y) = yx²

As we wanted. Lets find the endpoints of C

r(u) = u i + 2 u² j

r(0) = (0,0)

f(1) = (1,2)

Therefore,

$\int\limits_C {xy^2} \, dx + {yx^2} \, dy = f((1,2)) - f((0,0)) = \frac{1^2 * 2^2}{2} - \frac{0^20^2}{0} = \frac{4}{2} - 0 = 2$

The value of the line integral over C is 2.

Answer 2

After verifying the given vector field h is a gradient. The value of the line integral Over the endpoints of C is 2.

What is line integral?

A line integral is defined as an integral where the function to be integrated is evaluated along a curve.

${\displaystyle \int _{\mathcal {C}}f(\mathbf {r} )\,ds=\int _{a}^{b}f\left(\mathbf {r} (t)\right)\left|\mathbf {r} '(t)\right|\,dt.}$

To find such f, the first one over the variable x and the second one over the variable y.

$h(x,y) = xy^2 i + yx^2 j$

A general primitive of xy² i

$f_{1} (x,y) = x^2y^2/2 i + a(y)$

With a(y) a function that depends only on y.

$f_{2} (x,y) = x^2y^2/2 i + b(y)$

The function f(x,y) whose gradient is h is obtained by equating the expressions of f₁ and f₂.

f₁ and f₂ are equal when a(x) = b(x) = 0,

we get,

$f(x,y) = x^2y^2/2$

Here,

fx (x,y) = xy²

fy (x,y) = yx²

As we wanted. Lets find the endpoints of C

r(u) = u i + 2 u² j

r(0) = (0,0)

f(1) = (1,2)

Therefore,

$\int xy^{2} dx + yx^{2} dy\\f((1,2)- (0,0))\\1^2\times2^2/2 - 0\\= 4/2\\=2$

Thus, The value of the line integral over C is 2.

Learn more about line integral;

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