How Can I Simplify This Integral Using Green's Theorem?

Alexstrasza
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Homework Statement



I have a linear integral (e^xsiny-2)dx + (e^xcosy+x^2)dy
y≥0
2x=x^2+y^2

I used Green's theorem and got:

∬ (e^xcosy+2x) - (e^xcosy) dy dx
x bounds: from 0 to 2
y bounds: from 0 to sqrt(2x-x^2)

After solving all that stuff I get to:

∫ (2x) (sqrt(2x-x^2)) dx
x bounds: 0 to 2

I don't know how to calculate ∫ (2x) (sqrt(2x-x^2)) dx, I thought maybe I can turn the 2x into a derivative of 2x-x^2 but then I need to add +2 to the 2x and not sure if it is allowed?

I used an online integral calculator and it solved (answer: pi?) but the way of the solution was very very long so I think there must be an easier way to solve this (it is a test question).

Help please! ❀

Homework Equations



Calc II
 
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Alexstrasza said:

Homework Statement



I have a linear integral (e^xsiny-2)dx + (e^xcosy+x^2)dy
y≥0
2x=x^2+y^2

I used Green's theorem and got:

∬ (e^xcosy+2x) - (e^xcosy) dy dx
x bounds: from 0 to 2
y bounds: from 0 to sqrt(2x-x^2)

After solving all that stuff I get to:

∫ (2x) (sqrt(2x-x^2)) dx
x bounds: 0 to 2

I don't know how to calculate ∫ (2x) (sqrt(2x-x^2)) dx, I thought maybe I can turn the 2x into a derivative of 2x-x^2 but then I need to add +2 to the 2x and not sure if it is allowed?

I used an online integral calculator and it solved (answer: pi?) but the way of the solution was very very long so I think there must be an easier way to solve this (it is a test question).

Help please! ❀

Homework Equations



Calc II

Write ##2x = (2x-2) + 2##, to get
$$\int_0^2 2x \sqrt{2x-x^2} \: dx = \int_0^2 (2x-2) \sqrt{2x- x^2} \: dx + 2 \int_0^2 \sqrt{2x - x^2} \: dx.$$
The first integral is easy; the second one is also standard, and you can look it up or consult a textbook for the appropriate changes of variables.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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