Setup an integral for the curve

In summary, to find the length of the curve x = sqrt(64-y^2) with -4<=y<=4, we can set up an integral using the formula sqrt(1 + (y^2)/(64-y^2)) and then use substitution to solve for the length. Additionally, the graph of the curve is a circle with radius 8 centered at the origin.
  • #1
whatlifeforme
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Homework Statement


for the curve x= sqrt(64-y^2), -4<=y<=4

(identify from multiple choice
1. setup an integral for the curve.
2. identify the graph
3. find the length of the curve.

Homework Equations


x = sqrt(64-y^2), -4<=y<=4

The Attempt at a Solution


1. dx/dy = y*sqrt(64-y^2)
2. (dx/dy)^2 = y^2 * (64-y^2)
3. integral (-4 to 4) sqrt(1 + 64y^2 - y^4)

1. however, the answer for integral for the length is
L=Integral(-4,4) 8(64-y^2)^(-1/2) dy. I'm not sure how they got that?

2. I'm not sure how to graph this function.
 
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  • #2
hi whatlifeform! :smile:
whatlifeforme said:
x = sqrt(64-y^2), -4<=y<=4

1. dx/dy = y*sqrt(64-y^2)

no, you've correctly got the 2y and the 1/2,

but you haven't differentiated the function: the √ should become 1/2 1/√ :wink:

(generally, ()n in the chain rule becomes n()n-1)
 
  • #3
tiny-tim said:
hi whatlifeform! :smile:


no, you've correctly got the 2y and the 1/2,

but you haven't differentiated the function: the √ should become 1/2 1/√ :wink:

(generally, ()n in the chain rule becomes n()n-1)
what do you mean?
 
  • #4
i've got L = integral (-4,4) sqrt (1 + (y^2)/(64-y^2)) dy

however, I'm not sure how to get to: L=Integral(-4,4) 8(64-y^2)^(-1/2) dy
 
  • #5
update: got it solved. but I'm still not sure how to graph this function.
 
  • #6
whatlifeforme said:
update: got it solved. but I'm still not sure how to graph this function.
What do you get if you square both sides to get rid of the square root? Do you recognise the form that results?
 
  • #7
ie, what is x2 = 64 - y2 ? :smile:
 
  • #8
tiny-tim said:
ie, what is x2 = 64 - y2 ? :smile:

the graph of?
 
  • #9
x2 + y2 = a2

is an origin-centered circle with radius a
 

FAQ: Setup an integral for the curve

What is an integral?

An integral is a mathematical concept used to find the area under a curve, or the total accumulation of a function over a given interval.

How do you set up an integral for a curve?

To set up an integral for a curve, you need to evaluate the function at the limits of integration and subtract the two values. This will give you the area under the curve between those two points.

What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration, meaning you are finding the area under the curve between two given points. An indefinite integral does not have specific limits and represents the antiderivative of a function.

When do you use the definite integral?

The definite integral is used when you need to find the exact area under a curve between two specific points, or when calculating the total amount of something over a given interval.

What are some common methods for solving integrals?

Some common methods for solving integrals include using the fundamental theorem of calculus, substitution, integration by parts, and trigonometric substitution.

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