Let F = xyi - zk and S be the cone z = sqrt(x^2 + y^2), 0 <= z <= 1. Evaluate the surface integral ∬_S F · n dS oriented outward (i.e., away from the z-axis).
Added by Cheryl R.
Step 1
The surface S is given by the equation z = sqrt(x^2 + y^2), which can be rewritten as z^2 = x^2 + y^2. This is the equation of a cone with its vertex at the origin and opening along the z-axis. The gradient of the surface S is given by ∇z = <∂z/∂x, ∂z/∂y, -1>. Show more…
Show all steps
Close
Your feedback will help us improve your experience
Uma Kumari and 64 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 surface integral $$ \iint_{\sigma} f(x, y, z) d S $$ $f(x, y, z)=z^{2} ; \sigma$ is the portion of the cone $z=\sqrt{x^{2}+y^{2}}$ between the planes $z=1$ and $z=2$
TOPICS IN VECTOR CALCULUS
Surface Integrals
Find the surface integral of F(x,y,z) = x^2 over the cone z = sqrt(x^2 + y^2), 0 ≤ z ≤ 1.
Israel H.
Evaluate the surface integral. $$\begin{array}{l}{\iint_{S} x^{2} z^{2} d S} \\ {S \text { is the part of the cone } z^{2}=x^{2}+y^{2} \text { that lies between the }} \\ {\text { planes } z=1 \text { and } z=3}\end{array}$$
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