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# Another maths question watch

1. for part d, narb..

You integrated between π/4 and π/2 for your values of theta.

Surely for the area within the ellipse between P and the point where the ellipse touches the x-axis i.e: (x,y) (5,0).. you need to integrate between theta=0 and theta=π/4 .. right? Because when y=0, theta=0

so first if you integrate under the ellipse between @=0 and @=π/4

= INT [y.(dx/[email protected])[email protected]]

= INT [([email protected])([email protected])[email protected]]

= INT [-20.([email protected])^[email protected]]

= -10. INT [1 - [email protected]@]

= -10.[@ - 1/2 (sin [email protected])]

= -10.{[0 - 1/2 (sin 0)] - [π/4 - 1/2 (sin(π/2)]}

= (5π/2 - 5) units^2

________________

and then find the area of the triangle PQR, where Q ((5/2)root2 , 0)..

= [(5root2) - (5/2)root2]. root2

= 10 - 5 = 5 units^2

_______

= 5 - (5π/2 - 5) = (10 - 5π/2) units^2

anyone agree?! (ppppplllllllease?!)
2. I seem to be getting a different answer to everyone else....I integrated the curve between x=5√2 and x=5 (i.e. from the point P to where the curve crosses the x-axis). So in terms of 'θ' the integration is from pi/4 to 0
So, in the end my answer was: 0.5 (20 - 5pi)
3. (Original post by Sahir)
for part d, narb..

You integrated between π/4 and π/2 for your values of theta.

Surely for the area within the ellipse between P and the point where the ellipse touches the x-axis i.e: (x,y) (5,0).. you need to integrate between theta=0 and theta=π/4 .. right? Because when y=0, theta=0

so first if you integrate under the ellipse between @=0 and @=π/4

= INT [y.(dx/[email protected])[email protected]]

= INT [([email protected])([email protected])[email protected]]

= INT [-20.([email protected])^[email protected]]

= -10. INT [1 - [email protected]@]

= -10.[@ - 1/2 (sin [email protected])]

= -10.{[0 - 1/2 (sin 0)] - [π/4 - 1/2 (sin(π/2)]}

= (5π/2 - 5) units^2

________________

and then find the area of the triangle PQR, where Q ((5/2)root2 , 0)..

= [(5root2) - (5/2)root2]. root2

= 10 - 5 = 5 units^2

_______

= 5 - (5π/2 - 5) = (10 - 5π/2) units^2

anyone agree?! (ppppplllllllease?!)
4. (Original post by mockel)
yay, good stuff that was well tough!

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