Integrate along curve, book has wrong answer?

Click For Summary
SUMMARY

The discussion centers on the integration of a half-circle defined by the equation \(y = a \sin(t)\) over the interval from 0 to \(\pi\). The integral \(\int_0^{\pi} a \sin(t) dt\) evaluates to \(2a\), while the book incorrectly states the answer as \(2a^2\). A key point raised is the importance of using the differential arc length \(ds\) in the integral, which is calculated as \(ds = \sqrt{(dx/dt)^2 + (dy/dt)^2} dt\). This clarification resolves the discrepancy between the calculated and book answers.

PREREQUISITES
  • Understanding of parametric equations, specifically \(y = a \sin(t)\)
  • Knowledge of arc length calculation in calculus
  • Familiarity with definite integrals and their evaluation
  • Basic trigonometric identities and properties
NEXT STEPS
  • Study the derivation of arc length formulas in calculus
  • Learn about parametric equations and their applications in integration
  • Explore the concept of differential elements in integrals
  • Review common mistakes in integral calculus, particularly in geometric contexts
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques, as well as educators looking to clarify common misconceptions in integral calculus involving parametric curves.

Addez123
Messages
199
Reaction score
21
Homework Statement
$$\int _C y ds$$
where C is determined by
$$x^2+y^2=a^2, y >= 0$$
Relevant Equations
Math
So it's basically a half circle with radius a.
y = asin(t)
$$\int_0^{\pi} asin(t) dt = -acos(t) |_0^{\pi} = 2a$$

The book says the answer is ##2a^2##, but maybe that's wrong?
 
Physics news on Phys.org
Addez123 said:
Homework Statement:: $$\int _C y ds$$
where C is determined by
$$x^2+y^2=a^2, y >= 0$$
Addez123 said:
So it's basically a half circle with radius a.
y = asin(t)
##\int_0^{\pi} asin(t) dt = -acos(t) |_0^{\pi} = 2a##
The book says the answer is ##2a^2##, but maybe that's wrong?
I believe the book's answer. Notice that the integral includes ds, not dt.
Note that ##ds = \sqrt{(dx/dt)^2 + (dy/dt)^2}dt##
 
Last edited:
  • Like
Likes   Reactions: vanhees71, Addez123 and FactChecker
Thanks! Calculating dS was what I was missing, now it checks out!
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
12
Views
2K
Replies
23
Views
2K