Finding the Area of an Ellipse in Calculus III

In summary, the problem is to find the area of the ellipse (x^2)/4 + (y^2)/9 = 1 using the equation (1/2) * Integral of xdy - ydx = area of a Region. The suggested approach is to solve for y and then integrate from x = 0 to x = 2, which will give 1/4 of the area inside the ellipse. Another approach is to use double integrals to evaluate \iint dA for the region.
  • #1
shinobi12
16
0

Homework Statement



Find the area of the ellipse (x^2)/4 + (y^2)/9 = 1

Homework Equations



(1/2) * Integral of xdy - ydx = area of a Region


The Attempt at a Solution



The problem is weird to me because its complicated to attempt it in rectangular or polar coordinates)
 
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  • #2
How is the equation you give in #2 relevant to this problem? It's possible that it is, but if so, you'll need to refresh my memory as to why that's so. Also, you'll need a definite integral, with limits of integration.

The way I would approach this problem is to solve for y (the positive solution) and integrate from x = 0 to x = 2. That number would be 1/4 of the area inside the ellipse.
 
  • #3
Well since the title of the thread is calc 3, I guess this means you have learned about double integrals already. Evaluate [tex]\iint dA[/tex] for that region to get the area.
 

1. What is the formula for finding the area in Calculus III?

The formula for finding the area in Calculus III is the definite integral. It is represented by ∫f(x)dx, where f(x) is the function and dx is the infinitesimal change in the variable of integration.

2. How is finding the area in Calculus III different from Calculus I and II?

Finding the area in Calculus III involves integration in multiple dimensions, whereas Calculus I and II typically only deal with single-variable integration. This means that the area is calculated over a region in space rather than just on a single curve.

3. What are the different methods for finding the area in Calculus III?

The main methods for finding the area in Calculus III include iterated integrals, double integrals, and triple integrals. These methods differ in the number of variables and dimensions they involve, but all ultimately use the same concept of integrating a function over a region in space.

4. Can Calculus III be used to find the volume of a solid?

Yes, Calculus III can be used to find the volume of a solid. This is done by integrating the cross-sectional area of the solid with respect to the variable of integration. This method is commonly known as the method of cylindrical shells.

5. How can finding the area in Calculus III be applied in real life?

Finding the area in Calculus III has many practical applications in fields such as engineering, physics, and economics. It can be used to calculate the volume of a tank or the surface area of a 3D object, as well as to solve optimization problems involving multiple variables.

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