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# Integration with parametic equation Watch

1. I need help please on question 15)d)
Attached Images
2. Scan0223.pdf (343.6 KB, 133 views)
3. (Original post by otrivine)
I need help please on question 15)d)
What have you done so far? I'd start by finding the area of the right-angled triangle you've drawn on the figure and subtract from it the area under the curve.
4. (Original post by davros)
What have you done so far? I'd start by finding the area of the right-angled triangle you've drawn on the figure and subtract from it the area under the curve.
I integrated 4sinx which is -4cosx+c ? assume x is detha,

But I dont get the limits?
5. (Original post by otrivine)
I need help please on question 15)d)
Firstly, it's unclear from the diagram but I think that the required area is just the lower bit that looks like a triangle - not the upper bit with the ink lines on.
There's a neat way and a hackish way. You can find y in terms of x, by finding theta in terms of x and substituting. Alternatively, and much nicer, you can use the formula:
.
6. (Original post by Smaug123)
Firstly, it's unclear from the diagram but I think that the required area is just the lower bit that looks like a triangle - not the upper bit with the ink lines on.
There's a neat way and a hackish way. You can find y in terms of x, by finding theta in terms of x and substituting. Alternatively, and much nicer, you can use the formula:
.
No I used the area formula
7. (Original post by otrivine)
No I used the area formula
Ah, if there's a formula that spits out what you need then feel free to ignore me
8. (Original post by Smaug123)
Ah, if there's a formula that spits out what you need then feel free to ignore me
But dont get how they did the limit?
9. (Original post by otrivine)
No I used the area formula
What area formula

Smaug is correct

You found the limits when you found the co-ordinates of P and R
10. Why don't you create a triangle with the first quadrant inside it. Then you can take away the area of the triangle away from the area of the curve from the triangle.
11. (Original post by TenOfThem)
What area formula

Smaug is correct

You found the limits when you found the co-ordinates of P and R
Because to find the area for intgration , you integrate y with the limits
12. (Original post by otrivine)
Because to find the area for intgration , you integrate y with the limits
Yes, as Smaug said
13. (Original post by TenOfThem)
Yes, as Smaug said
and so you integrate only 4sinx
14. (Original post by otrivine)
Because to find the area for intgration , you integrate y with the limits
You integrate y(x) with respect to the x-limits. If you integrate with respect to a parameter then you must ensure that your limits are the appropriate values of the parameter.
15. (Original post by otrivine)
Because to find the area for intgration , you integrate y with the limits
Can you give us the expression you're integrating, so we know what you mean by "area formula"? Then we might be able to help with the limits…
16. (Original post by otrivine)
and so you integrate only 4sinx
You need to integrate y with respect to x

but y = 2Sin(theta) so you need to do what Smaug said to do
17. this is what I mean?
Attached Images
18. Scan0224.pdf (242.7 KB, 73 views)
19. (Original post by otrivine)
this is what I mean?
The limits are the theta limits, and you need to rewrite your dx in terms of d(theta) as smaug showed you earlier.
20. (Original post by otrivine)
this is what I mean?
That is incorrect

You can not integrate Sin(theta) with respect to x

You need to use the formula that Smaug showed you

The limits are

x/theta at point P
x/theta when y=0
21. (Original post by otrivine)
this is what I mean?
You haven't actually integrated correctly. You've integrated d(theta) instead of d(x).
22. (Original post by Smaug123)
You haven't actually integrated correctly. You've integrated d(theta) instead of d(x).
ok

what would you integrate then can you show me

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