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Among all smooth, simple closed curves in the plane, oriented counterclockwise, find the one along which the work done by $$\mathbf{F}=\left(\frac{1}{4} x^{2} y+\frac{1}{3} y^{3}\right) \mathbf{i}+x \mathbf{j}$$ is greatest. (Hint: Where is (curl $\mathbf{F}$ ) $\cdot \mathbf{k}$ positive?)

Calculus 3

Chapter 16

Integrals and Vector Fields

Section 4

Green’s Theorem in the Plane

Vector Functions

Missouri State University

Campbell University

University of Michigan - Ann Arbor

University of Nottingham

Lectures

03:04

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x. The input of a function is called the argument and the output is called the value. The set of all permitted inputs is called the domain of the function. Similarly, the set of all permissible outputs is called the codomain. The most common symbols used to represent functions in mathematics are f and g. The set of all possible values of a function is called the image of the function, while the set of all functions from a set "A" to a set "B" is called the set of "B"-valued functions or the function space "B"["A"].

08:32

In mathematics, vector calculus is an important part of differential geometry, together with differential topology and differential geometry. It is also a tool used in many parts of physics. It is a collection of techniques to describe and study the properties of vector fields. It is a broad and deep subject that involves many different mathematical techniques.

03:58

Maximizing work Among all …

10:55

00:47

Evaluate $\int \frac{x d x…

Okay problem 44 point: we want to calculate the work, so here's f- and the only thing we want to know is what the curve is right and just use a green serum and see so the work should be double integral. I don't know what this region is integral, so here's partial part x, which is 1 and minus partial part, is 1 minus 1 over 4 x square and minus y square right. Okay, here this- and we want this to be- and we want to make this integral. As great as possible on so when this these terms say f, x, y but x, y is greater than 0 integral should be positive, and if this is smaller than 0, and that should be not not good because it makes the smaller right. So the integral should be where d should be exactly where f is greater than 0. So, let's start 1 over 4 x square plus 2 is smaller than 1. Is a lives right, so here's a 2 heresies. I don't care. I don't even think i care about this x is equals to 2 cosine, theta and y to be 2 r, cosine, theta and y to be our sine theta. So partial x, partial r should be 2 cosine theta and the jackal matric minus 2 r si theta sine theta, of course, minus minus. There'S no minus here, just our cosine theta. So the absolute value of the jocko determination should be 2 r right. So this equals to this 1 minus r square and times 2 r d, r d, t and r from just 0 to to 1 right are from 0 to 1. Pay here is our integral just calculate so: that's 2 r, so 1 minus 2 r cube. So, oh, we haven't got the seda from 0 to 2 pi. I thought about that. So 1 minus 1 over 2 pi times 2 pi. So the answer should be piimaximal force right.

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