Question

Evaluate the iterated integral.
π/4 cosθ
∫ ∫ sinθdrdθ
0 0

Explain please

Answer 1

are you missing an r from that integrand/

Answer 2

no thats the entire problem

Answer 3

\[\text{but the volume element should be } "r\text dr \text d \theta"\]

Answer 4

This is how the problem is in the textbook I have to take the inner integral part first and then the outside one

Answer 5

I believe the equation is correct as is. drdtheta gives the order to integrate in, does it not?

Answer 6

yes

Answer 7

ok

Answer 8

This should clear up any confusion... http://tutorial.math.lamar.edu/Classes/CalcIII/IteratedIntegrals.aspx

Answer 9

i looked at that site before but it's not helpful to me

Answer 10

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.

Answer 11

\[ \int\limits_0^{π/4} \int\limits_0^{cos \theta} \sin \theta \text{d} r \text{d} \theta\]
\[=\int\limits_0^{π/4} r\sin\theta \text d r |_0^{cos\theta} \text d\theta \]
\[=\int\limits_0^{π/4}cos{\theta}\sin{\theta}\text d \theta\]

Answer 12

Yep, that's it ^

Answer 13

\[=\int\limits_{0}^{π/4}\frac{\sin(2\theta)}{2}\text d {\theta} \]

Answer 14

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)

Answer 15

\[=\frac{1}{4} \cos {(2 \theta)} |_{0}^{π/4} \]

Answer 16

The derivative of sinx=cosx, so you can use U substitution, thus it should be sin^2(x)/2

Answer 17

\[=\frac{1}{4}\left(cos(π/2) -\cos(0)\right)\]

Answer 18

=1/4

Answer 19

thanks so much i understood it!

Answer 20

i have used this trig identity \[\sin(2x) = 2 \sin ( x) \cos (x)\]
and
remember the integral of sin(2x) is cos(2x)/2

Answer 21

thank you! I'll practice u-sub more and memorizing trig identities.

Answer 22

i didn't use any substitution

Answer 23

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 ?

Answer 24

\[\cos(\theta) \sin(\theta)=\sin(2\theta)\]
\[\int \sin(\theta) \text d\theta=\cos(\theta)+c\]
\[\int \sin(2\theta) \text d\theta=\frac{\cos(\theta)+c}{2}\]

Answer 25

But, sin(2θ) = 2sin(x)cos(x). You don't have a 2 in front.

Answer 26

\[\sin(2θ) = 2\sin(x)\cos(x)\]\[\qquad \downarrow\]\[\frac{\sin(2θ)}{2} = \sin(x)\cos(x)\]


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

Answer 27

Ah yes. Thanks