Evaluate the line integral, where C is the given curve: $$ \int_C y^3 ds, C: x = t^3, y = t, 0 \le t \le 4 $$ ScalceT9 16.XP.2.001.
Added by Michael R.
Close
Step 1
First, we need to find $$ ds $$. For a parametric curve given by $$ x = x(t) $$ and $$ y = y(t) $$, $$ ds $$ is given by the formula: $$ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt $$ Step 2: Calculate the derivatives of x and y Show more…
Show all steps
Your feedback will help us improve your experience
Adi S and 61 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Evaluate the line integral, where C is the given curve. C (x/y) ds, C: x = t3, y = t4, 1 ≤ t ≤ 4
Adi S.
Evaluate the line integral, where $C$ is the given curve. $$\int_{C} y^{3} d s, \quad C : x=t^{3}, y=t, 0 \leqslant t \leqslant 2$$
VECTOR CALCULUS
Line Integrals
$1-16$ Evaluate the line integral, where $C$ is the given curve. $$\int_{C} y^{3} d s, \quad C : x=t^{3}, y=t, 0 \leqslant t \leqslant 2$$
Vector Calculus
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD