need help with this calc qn

[willow hemlock]

i dont really know how to approach this qn i think you have to find the equation of the circle but idk how to even find the center of the circle

[drifting musk]

Can you find just the x-coordinate of the centre?

With that, there are several ways to find the centre, but the easiest is to remember that any line tangent to the circle is perpendicular to the connecting radius. And at the point of intersection you're given, the tangent to the parabola is tangent to the circle.

[willow hemlock]

ohh so is the center of the circle (2.16,1.04)

then the radius is 1.04 right

[drifting musk]

Yup

[willow hemlock]

so the equation of the circle is (x-2.16)^2+(y-1.04)^2=1.04^2 right

do i make y the subject and integrate that

[drifting musk]

Oh yeah, integrating over y is a good idea

[willow hemlock]

ok i think im stuck

idk how to integrate the last part

[drifting musk]

I thought you were planning to integrate from y=0 to y=1.44 ...dy

Integrals of the type that you set up there are done with trig substitution

[willow hemlock]

whats trig substitution

[willow hemlock]

wdym

[drifting musk]

It's a form of u-substitution that makes use of the pythagorean identity. Your best bet is to look up trig substitution on youtube or kahn academy

[willow hemlock]

i dont think i have learnt that yet

[drifting musk]

You can either integrate along y or along x. If you integrate along x, then you're going to need 3 separate integrals for the different regions of the shaded area

[willow hemlock]

but if i integrate on y i will still have a square root that i cant integrate right

[drifting musk]

You can figure out the part that's underneath the circle by finding the area of a square and subtracting a quarter of a circle, but for the shaded part that's above the circle, you're going to need trig substutition

[drifting musk]

yes

[willow hemlock]

so is this qn not possible

[drifting musk]

Trig substitution is really just an application of u-substitution, but if you've never seen it before then I'm not sure how you're supposed to do this question

[willow hemlock]

damn

[untold wedge]

I'm not 100% sure, but this is what I came up with

[lost finch]

animeonfire

[untold wedge]

I hope this helps, though it is a mess

[drifting musk]

integrating x^2 is easy, and it doesn't give a square root, but it wouldn't even matter if it did

Not a bad approach overall though. Definitely harder than the regular integral, but 🤷‍♂️

Do you have a plan for finding the segment of the circle?

[slow light]

it's simpler than you think

the answer is around 0.74764299

So, notice the area is simply the area of the rectangle in cyan minus the two other areas

The area of the rectangle is 3.1104

First, integrate the area in blue. Recognize that this area is the same area as the integral of (x)^1/2 from 0 to 1.44.

Therefore, the area of the blue is 1.152

So, 3.1104 - 1.152 = 1.9584

Now, let's find the green area. Also recognize that this green area is the same area as (1-x^2)^1/2 from -1 to 0.44

Therefore, this area is 1.21075701

So, subtract the previous area.

1.9584 - 1.21075701 = 0.74764299

Hope this helped

@drifting musk @willow hemlock @untold wedge

[slow light]

Tldr it's basically taking the area of the bigger rectangle and subtracting the non shaded areas

[drifting musk]

Okay, but how did you get this? That's the entire point. We already knew everything you said, although you did put it with a nice picture. But you didn't actually help with the part that wasn't finished here.

[slow light]

So basically we know the circle has a radius 1, and since the area is only in half of the circle, we can try modeling it with the graph of y = (1-x^2)^1/2

And the y coord on the problem is 1.44

[drifting musk]

stop

[drifting musk]

this is obvious. I know this.

I mean how did you evaluate the integral

[slow light]

Oh

U use trig sub

[drifting musk]

Because I already said it's easy to do with trig substitution, but OP doesn't even know substitution

[slow light]

Oh

Shit

Well then idk

Idk how else to find the circles area then

[drifting musk]

I don't think it's possible to find a segment of a circle like this without trig-substitution (or some replacement for it, like a really lucky guess-and-check)

[slow light]

Yea

Ig u can just enter it into mathway if u can't solve it

💀💀

[willow hemlock]

Wait how do you know the circle has radius 1

ok i found the solution my teacher gave us

I dont really get it though

@drifting musk

[drifting musk]

It doesn't exactly, but when you're figuring out what to do you don't care what the actual radius is because it's easy to scale the areas accordingly.

[drifting musk]

It's a slightly different approach where you connect the two tangent points with a straight line, find the area under that, then remove the segment by knowing what angle those to tangent points make when joined to the center of the circle. What specifically do you not get?

[willow hemlock]

oh

[slow light]

Yeah this method probably more precise

[slow light]

It's close enough u can just give some confidence interval or something to account for the real radius

There's a lot of nuance