Setup an integral for the curve

[whatlifeforme]

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.

 

[tiny-tim]

hi whatlifeform!

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/√

(generally, ()ⁿ in the chain rule becomes n()^n-1)

 

[whatlifeforme]

hi whatlifeform!


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

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

(generally, ()ⁿ in the chain rule becomes n()^n-1)

what do you mean?

 

[whatlifeforme]

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

 

[whatlifeforme]

update: got it solved. but I'm still not sure how to graph this function.

 

[haruspex]

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?

 

[tiny-tim]

ie, what is x² = 64 - y² ?

 

[whatlifeforme]

ie, what is x² = 64 - y² ?

the graph of?

 

[dx]

x² + y² = a²

is an origin-centered circle with radius a