anonymous
  • anonymous
Evaluate the iterated integral. π/4 cosθ ∫ ∫ sinθdrdθ 0 0 Explain please
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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UnkleRhaukus
  • UnkleRhaukus
are you missing an r from that integrand/
anonymous
  • anonymous
no thats the entire problem
UnkleRhaukus
  • UnkleRhaukus
\[\text{but the volume element should be } "r\text dr \text d \theta"\]

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anonymous
  • anonymous
This is how the problem is in the textbook I have to take the inner integral part first and then the outside one
anonymous
  • anonymous
I believe the equation is correct as is. drdtheta gives the order to integrate in, does it not?
anonymous
  • anonymous
yes
UnkleRhaukus
  • UnkleRhaukus
ok
anonymous
  • anonymous
This should clear up any confusion... http://tutorial.math.lamar.edu/Classes/CalcIII/IteratedIntegrals.aspx
anonymous
  • anonymous
i looked at that site before but it's not helpful to me
anonymous
  • anonymous
You just integrate sinθ in regards to dr and as there is no dr don't you treat dr as 1 and the int of 1 = x, then you evaluate on the end points and proceed to take the second integral in regards to theta within the second region.
UnkleRhaukus
  • UnkleRhaukus
\[ \int\limits_0^{π/4} \int\limits_0^{cos \theta} \sin \theta \text{d} r \text{d} \theta\] \[=\int\limits_0^{π/4} r\sin\theta \text d r |_0^{cos\theta} \text d\theta \] \[=\int\limits_0^{π/4}cos{\theta}\sin{\theta}\text d \theta\]
anonymous
  • anonymous
Yep, that's it ^
UnkleRhaukus
  • UnkleRhaukus
\[=\int\limits_{0}^{π/4}\frac{\sin(2\theta)}{2}\text d {\theta} \]
anonymous
  • anonymous
the answer is 1/4 but im getting 1/2? i understood everything I was just stuck on where to go from cosxsinx (the antiderivative)
UnkleRhaukus
  • UnkleRhaukus
\[=\frac{1}{4} \cos {(2 \theta)} |_{0}^{π/4} \]
anonymous
  • anonymous
The derivative of sinx=cosx, so you can use U substitution, thus it should be sin^2(x)/2
UnkleRhaukus
  • UnkleRhaukus
\[=\frac{1}{4}\left(cos(π/2) -\cos(0)\right)\]
UnkleRhaukus
  • UnkleRhaukus
=1/4
anonymous
  • anonymous
thanks so much i understood it!
UnkleRhaukus
  • UnkleRhaukus
i have used this trig identity \[\sin(2x) = 2 \sin ( x) \cos (x)\] and remember the integral of sin(2x) is cos(2x)/2
anonymous
  • anonymous
thank you! I'll practice u-sub more and memorizing trig identities.
UnkleRhaukus
  • UnkleRhaukus
i didn't use any substitution
anonymous
  • anonymous
I'm confused how you went from cosθsinθdθ to sin(2θ)/2dθ. You can use u-sub with cosθsinθ because the derivative of sinx=cosx, so it should be Sinx^2(x)/2, how did you get sin(2θ)/2 ?
UnkleRhaukus
  • UnkleRhaukus
\[\cos(\theta) \sin(\theta)=\sin(2\theta)\] \[\int \sin(\theta) \text d\theta=\cos(\theta)+c\] \[\int \sin(2\theta) \text d\theta=\frac{\cos(\theta)+c}{2}\]
anonymous
  • anonymous
But, sin(2θ) = 2sin(x)cos(x). You don't have a 2 in front.
UnkleRhaukus
  • UnkleRhaukus
\[\sin(2θ) = 2\sin(x)\cos(x)\]\[\qquad \downarrow\]\[\frac{\sin(2θ)}{2} = \sin(x)\cos(x)\] remember that sine is really \[\cos\theta=\frac{e^{i\theta}+e^{-i\theta} }{2}\] and integrating will not affect the argument of the trigonometric function
anonymous
  • anonymous
Ah yes. Thanks

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