Use of integration to find area

In summary, the area enclosed by the curve x = t^2 -2t, y = t^0.5, and the y-axis can be found by setting x = 0 and solving for t, giving the limits of integration. Then, you can use the given equation, ∫g(t) f'(t) dt, or modify it to ∫f(t) g'(t) dt to find the area. Both equations will give you the same result, but one will be the negative of the other.
  • #1
Calpalned
297
6

Homework Statement


Find the area enclosed by the curve x = t^2 -2t, y = t^0.5 and the y axis

Homework Equations


Area of a parametric curve = ∫g(t) f'(t) dt, where g(t) = y and f(t) = x

The Attempt at a Solution


I believe that the limits of integration by be found by setting x and y equal to each other and solving for y. Do I still use the given equation, or do I have to modify it as this question asks about the y-axis, not the x-axis.

Thanks
 
Physics news on Phys.org
  • #2
Calpalned said:

Homework Statement


Find the area enclosed by the curve x = t^2 -2t, y = t^0.5 and the y axis

Homework Equations


Area of a parametric curve = ∫g(t) f'(t) dt, where g(t) = y and f(t) = x

The Attempt at a Solution


I believe that the limits of integration by be found by setting x and y equal to each other and solving for y. Do I still use the given equation, or do I have to modify it as this question asks about the y-axis, not the x-axis.

Thanks

The limits of integration are the values of t where your curve intersects the y-axis. The y-axis is where the value of x equals 0. Solve that.
 
  • #3
I solved t^2 - 2t = x = 0 and I got t = 2 and t = 0, so those are my limits
Now, do I still use ∫g(t) f'(t) dt or do I have to modify the equation to ∫f(t) g'(t) dt ?
 
  • #4
Calpalned said:
I solved t^2 - 2t = x = 0 and I got t = 2 and t = 0, so those are my limits
Now, do I still use ∫g(t) f'(t) dt or do I have to modify the equation to ∫f(t) g'(t) dt ?

You can actually use either one. One will give you the negative of the other. You can see this by looking at integration by parts.
 

1. What is integration and how is it used to find area?

Integration is a mathematical concept that involves finding the area under a curve. It is used to find the exact value of an area that cannot be easily calculated using basic geometric formulas.

2. What are the different types of integration methods that can be used to find area?

The two main types of integration methods used to find area are the definite integral and the indefinite integral. The definite integral calculates the area between two specific points on a curve, while the indefinite integral calculates the general area under a curve.

3. How does the Riemann sum method relate to integration and finding area?

The Riemann sum method is a technique used to approximate the area under a curve by dividing the area into smaller, simpler shapes. This method is used as the basis for integration, where the number of smaller shapes used approaches infinity, resulting in a more accurate calculation of the area.

4. Can integration be used to find the area of irregular shapes?

Yes, integration can be used to find the area of irregular shapes. By dividing the irregular shape into smaller, simpler shapes, the area can be approximated using the Riemann sum method. As the number of smaller shapes approaches infinity, the approximation becomes more accurate.

5. What are some real-world applications of using integration to find area?

Integration is used in various fields, such as physics, engineering, and economics, to calculate real-world quantities such as displacement, work, and revenue. For example, in physics, integration is used to calculate the area under a velocity-time graph to determine the total distance traveled by an object.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
160
  • Calculus and Beyond Homework Help
Replies
8
Views
600
  • Calculus and Beyond Homework Help
Replies
3
Views
571
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
991
  • Calculus and Beyond Homework Help
Replies
2
Views
391
  • Calculus and Beyond Homework Help
Replies
24
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
471
  • Calculus and Beyond Homework Help
Replies
3
Views
287
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top