Question
Find the area of the given surface.The portion of the cone $z=\sqrt{x^{2}+y^{2}}$ that lies inside the cylinder $x^{2}+y^{2}=2 x$
Step 1
The equation of the cone $z=\sqrt{x^{2}+y^{2}}$ becomes $z=r$ and the equation of the cylinder $x^{2}+y^{2}=2x$ becomes $r^{2}=2r\cos\theta$ or $r=2\cos\theta$. Show more…
Show all steps
Your feedback will help us improve your experience
Mohamed Raafat Mohamed and 51 other Calculus 3 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
Find the area of the surface. The portion of the cone $z=2 \sqrt{x^{2}+y^{2}}$ inside the cylinder $x^{2}+y^{2}=4$
Multiple Integration
Surface Area
Find the area of the part of the cylinder $x^{2}+y^{2}=2 a y$ that lies outside the cone $z^{2}=x^{2}+y^{2}$
Vector Fields
Surfaces and Surface Integrals
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD