Question

Problem 9 (10 Marks) - Use surface integral to find the area of the part of the surface $z = x^2 + y^2$ that is below the plane $z = 4$ in the first octant. Note $\int \sqrt{a^2 + u^2} du = 0.5u\sqrt{a^2 + u^2} + 0.5a^2ln(u + \sqrt{a^2 + u^2})$.

Name: problem 9 10 marks use surface integral to find the area of the part of the surface z x2 y2 that is below the plane z 4 in the first octant note int sqrta2 u2 du 05usqrta2 u2 05a2lnu sqrta2 11738
Uploaded: 2020-03-23T14:58:11-08:00
Duration: 9 min 43 s
Channel: Carlos Pinilla
Description: problem 9 10 marks use surface integral to find the area of the part of the surface z x2 y2 that is below the plane z 4 in the first octant note int sqrta2 u2 du 05usqrta2 u2 05a2lnu sqrta2 11738

          Problem 9 (10 Marks) - Use surface integral to find the area of the part of the surface $z = x^2 + y^2$ that is below the plane $z = 4$ in the first octant.
Note $\int \sqrt{a^2 + u^2} du = 0.5u\sqrt{a^2 + u^2} + 0.5a^2ln(u + \sqrt{a^2 + u^2})$.

problem 9 10 marks use surface integral to find the area of the part of the surface z x2 y2 that is below the plane z 4 in the first octant note int sqrta2 u2 du 05usqrta2 u2 05a2lnu sqrta2 11738

Added by Christine B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Carlos Pinilla

03/23/2020

Step 1

The surface is given by $z = x^2 + y^2$. We want to find the area of the part of the surface below the plane $z = 4$ in the first octant. The surface area is given by the surface integral: $$A = \iint_S dS$$ where $dS = \sqrt{1 + (\frac{\partial z}{\partial x})^2 Show more…

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Problem 9 (10 Marks) - Use surface integral to find the area of the part of the surface $z = x^2 + y^2$ that is below the plane $z = 4$ in the first octant. Note $\int \sqrt{a^2 + u^2} du = 0.5u\sqrt{a^2 + u^2} + 0.5a^2ln(u + \sqrt{a^2 + u^2})$.