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Find an equation for the surface obtained by rotating the curve $ y = \sqrt{x} $ about the x-axis.

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01:07

Wen Zheng

01:28

Carson Merrill

Calculus 3

Chapter 12

Vectors and the Geometry of Space

Section 6

Cylinders and Quadric Surfaces

Vectors

Campbell University

Oregon State University

University of Michigan - Ann Arbor

Boston College

Lectures

02:56

In mathematics, a vector (from the Latin word "vehere" meaning "to carry") is a geometric entity that has magnitude (or length) and direction. Vectors can be added to other vectors according to vector algebra. Vectors play an important role in physics, engineering, and mathematics.

11:08

In mathematics, a vector (from the Latin word "vehere" which means "to carry") is a geometric object that has a magnitude (or length) and direction. A vector can be thought of as an arrow in Euclidean space, drawn from the origin of the space to a point, and denoted by a letter. The magnitude of the vector is the distance from the origin to the point, and the direction is the angle between the direction of the vector and the axis, measured counterclockwise.

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Find an equation for the s…

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All right, we want to take the curves, Y equals root X and rotate it um around the X axis. Let me try rather one more time. See if I can draw better. Okay, that's a little bit better. So what's happening is it's going to come out of the page and then back into the page and you're going to get this nice um Carrabba, Lloyd looking thing. Okay, so what happens? Uh let's first ask ourselves if that equation is still valid. Um So now we have X. Y and then Z is actually coming out of the page. So this is a funny looking um three dimensional axis, we normally wouldn't drop like this um at every point on that circle is why still the root of X will definitely not, it's only that on the extremities. So okay, so we scrap that equation. That's annoying. We don't have anything anymore. So what do we have? Well, for each value of X, we have something like that red circle and that red circle has radius, route X and uh let's see uh the circle um so the X stays the same and y and Z both change right as we walk around that red thing. So um the circle is actually uh export dependent on y and Z. So we're going to get y squared plus Z squared and then it's going to equal the radius squared. Well, radius is route X. So you get root X squared. So um now the equation for the surface is y squared plus Z squared equals X. Um So this is a great problem is a lot of thinking. Um There's no like formulaic way to solve a problem like this, you've got to think about what's happening in three dimensions.

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