Integrate over area enclosed by curve

In summary: You just plugged in the value for dy/dx, which is correct. Your final answer is also correct, so you're good to go.
  • #1
sanitykey
93
0
Hi, I've been asked to use polar coordinates and to integrate

[tex]\int\int_{A}^{} x^2 + y^2 + \sqrt{x^2 + y^2}dxdy[/tex]

over the area A enclosed by the curve [tex]x^2 + y^2 = 4[/tex]

I know

[tex]x=rcos(\theta)[/tex]
[tex]y=rsin(\theta)[/tex]
[tex]dxdy=rdrd\theta[/tex]

So i think the integral can be written as

[tex]\int\int_{A}^{} r^3 \pm r^2 drd\theta[/tex]

and i think the limits might be:

[tex]\int_{\theta=0}^{2\pi}\int_{r=-2}^{2} r^3 \pm r^2 drd\theta[/tex]

But i think these are probably wrong and I'm not sure how to account for the negative areas and such because i can't use symmetry if it's a function like that can i?

If i do that i think the area comes out as [tex]64\pi/3[/tex]
 
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  • #2
r is never negative in polar coordintes. The lower limit of the r integral is zero. The theta variable is what takes you all the way around the curve. When a square root is written as you have been given, the positve square root is implied. The integrand is just r^3 + r.
 
  • #3
Ok i think i understand would that mean it should be [tex]\int_{\theta=0}^{2\pi}\int_{r=0}^{2} r^3 + r^2 drd\theta[/tex]


Which i think comes out with an answer of [tex]40\pi/3[/tex]

If the inetgrand is [tex]r^3 + r[/tex] does that mean the r from the [tex]rdrd\theta[/tex] is only multiplied to one of them? I wasn't sure when i was writing it and just guessed it'd be multiplied to both.

Thanks for your reply btw :)
 
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  • #4
sanitykey said:
Ok i think i understand would that mean it should be [tex]\int_{\theta=0}^{2\pi}\int_{r=0}^{2} r^3 + r^2 drd\theta[/tex]


Which i think comes out with an answer of [tex]40\pi/3[/tex]

If the inetgrand is [tex]r^3 + r[/tex] does that mean the r from the [tex]rdrd\theta[/tex] is only multiplied to one of them? I wasn't sure when i was writing it and just guessed it'd be multiplied to both.

Thanks for your reply btw :)
In any coordinates you use, dA multiplies the whole integrand. The r instead of r^2 was just from careless typing. Sorry. It should be r^2. I fixed it in my earlier post. Your latest expression looks good.
 
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  • #5
Thanks again my main problem with these types of questions is visualising what it actually means.
 
  • #6
Didn't think i should post this in a different thread as it's sort of the same topic but could someone please check this for me:

Question:

Integrate the integral below

[tex]\int\sqrt{1+x^2}ds[/tex]

along the curve [itex]y=\frac{x^2}{2}[/itex] between [tex]x=0[/tex] and [tex]x=1[/tex]. Remember that [itex]ds=\sqrt{dx^2 + dy^2}[/itex]

What i did:

[tex]\frac{dy}{dx}=x[/tex]


=> [tex]dy=xdx[/tex]


=> [tex]dy^2 = x^2 dx^2[/tex]


[tex]\int\sqrt{1+x^2}\sqrt{dx^2 + dy^2}[/tex]


=> [tex]\int\sqrt{1+x^2}\sqrt{dx^2 + x^2 dx^2}[/tex]


=> [tex]\int\sqrt{1+x^2}\sqrt{dx^2 (1+x^2)}[/tex]


=> [tex]\int\sqrt{1+x^2}\sqrt{1+x^2}dx[/tex]


=> [tex]\int1+x^2dx[/tex]


=> [tex]\int_{0}^{1}1+x^2dx[/tex]


=> [tex][1+\frac{1^3}{3}] = \frac{4}{3}[/tex]
 
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  • #7
That looks good.
 

Related to Integrate over area enclosed by curve

What is integration over area enclosed by a curve?

Integration over area enclosed by a curve is a mathematical process in which the total area bounded by a curve on a graph is calculated by using integration techniques. This technique is used in various fields of science, such as physics, engineering, and economics, to determine the total value or quantity of a given variable.

Why is integration over area enclosed by a curve important?

Integration over area enclosed by a curve is important because it allows us to find the total area bounded by a curve, which is crucial in many real-life applications. For example, in physics, integration over area can be used to calculate the work done by a force, and in economics, it can be used to find the total revenue or profit generated by a company.

How is integration over area enclosed by a curve calculated?

Integration over area enclosed by a curve is calculated by using the fundamental theorem of calculus. This involves breaking down the area into small rectangles and finding the sum of their areas, which is represented by the integral symbol. The integral is then evaluated using integration techniques, such as the substitution method or integration by parts.

What are some real-life examples of integration over area enclosed by a curve?

Integration over area enclosed by a curve has many practical applications. For instance, in physics, it can be used to calculate the displacement of an object, the amount of work done by a force, or the total energy of a system. In economics, it can be used to find the total revenue or profit generated by a company. It is also used in engineering to determine the total volume of a structure or the amount of material needed for a construction project.

What are some common challenges faced when integrating over area enclosed by a curve?

One of the main challenges of integration over area enclosed by a curve is choosing the correct limits of integration. This requires a deep understanding of the problem and the ability to visualize the curve on a graph. Another challenge is determining the correct integration technique to use, as different curves may require different methods. Additionally, dealing with complex curves or functions can also be a challenge in the integration process.

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