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Simple intergral using parametric equations

  • Thread starter Galadirith
  • Start date
  • #1
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Homework Statement



i. [tex]x = 3cost,[/tex]
ii. [tex]y = 9sin2t,[/tex]
iii. [tex]0\leq t < 2\pi[/tex]


iv.[tex]\int_0^\frac{\pi}{2} Asin2tsint \ dt[/tex]

2. The attempt at a solution

So this is what I am given and I am supposed to be able to show that this is the integral for the shadded area between the curve and the x-axis in the 1st quadrant (sorry I know that relys on a graph, but the question is related I have isnt related to the actually shadded area anyway), so I can show the the interval is correct and form the intergral using

[tex] \int y \frac{dx}{dt} dt [/tex]

but they state as answer that A is 27. Now I know that that must be the answer because the shaded region is above the x-axis, so the integral in those limits should be possitive, but whenever I try to from the intergral equation I get -27 becuase dx/dt is -3sint. Can anyone help me to see how I have gone wrong, thanks.
 

Answers and Replies

  • #2
109
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Ah I have figured out the answer, basically the curve cuts the x-axis at 0 and 3 (well at least the shaded region), and getting the paramater t for each value is x=0, t= pi/2 and x = 3, t=0. so in actuall fact on initial formualtion of the intergral I should get:

[tex] \int_{\frac{\pi}{2}}^0 -27sin2tsint \ dt [/tex]

and then if I reverse the limits of intergration of the intergral i get :

[tex] \int_0^{\frac{\pi}{2}} 27sin2tsint \ dt [/tex]

Thanks all who did look (or will) :-)
 

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