Find the exact area of the surface obtained by rotating the curve about the x-axis. x=1+2 y^2, 1

Find the exact area of the surface obtained by rotating the...

Key Concepts

Surface Area of Revolution

This concept refers to the process of generating a three-dimensional surface by rotating a two-dimensional curve about a specified axis. Calculating the surface area of such a surface involves setting up an integral that sums over the contributions of infinitesimal elements of the rotating curve, taking into account both the arc length and the path traveled during the rotation.

Arc Length Differential

The arc length differential, often denoted as ds, represents an infinitesimally small segment of the curve. It is expressed in terms of the derivatives of the curve's defining function. In surface area problems, ds is crucial because it is used in the integral to capture the element of the curve that, when rotated, contributes to the total surface area.

Definite Integration

Definite integration is the process of summing the contributions of infinitesimal elements over a specified interval to obtain a total quantity, such as area. In the context of surfaces of revolution, the integral accumulates the surface area elements, which are functions of the arc length differential and the distance from the axis of rotation, over the given bounds.

Parametrization of Curves

Parametrization involves expressing a curve in terms a parameter, which can simplify the computation of derivatives and integrals. This approach is particularly useful in rotations and surface area calculations, as it allows the curve to be described in a form that facilitates the application of the surface area formula when the curve is rotated about an axis.

Transcript

00:01 For this problem, we are asked to find the exact area of the surface obtained by rotating the curve x equals 1 plus 2 y squared for y between 1 and 2 about the x axis to start, we want to rearrange this to get y as a function of x y will be plus or minus the square root of x minus 1 over 2 just doing some simple rearranging subtract 1 from both sides by both sides by 2 take the root since we're rotating about the x axis here we only need to take the positive root so now having that we need to find y prime, the derivative, which we can find just by applying a chain rule here and simplifying down a little bit.

00:37 The result is going to be 1 over 2 root 2 times the square root of x minus 1, which means that our y prime squared, which will need for our surface area formula, is going to be 1 over 8 times x minus 1.

00:55 Now having that we can, last step we need to be careful about here.

01:00 We need to figure out what x of 1 the lower bound on y and x of 2 will be so x of 1 would be 1 plus 2 times 1 squared so just 1 plus 2 so that's going to be 3 then we'll have 1 plus 2 times 4 from 2 squared so 2 times 4 8 plus 1 is going to be 9 so now we have the surface area equation so we'll be integrating from 3 up to 9 of integrating from 3 up to 9 to 2 pi times y, which will be the square root of x minus 1 over 2, times the square root of 1 plus y prime squared.

01:41 So 1 plus 1 over 8 times x minus 1.

01:46 We are integrating over x.

01:48 So the first thing that we can do is do a little bit of rearranging and simplification.

01:53 So we can recognize that root x minus 1 over 2 is the same thing as root x minus 1 over root 2.

01:58 So we have this 2, which, you know, that would be 2 to the power of 1.

02:02 Then we're dividing it by 2 to the power of 1 half.

02:05 So we'll end up with a 2 to the power of 1 half on top, or a root 2 in the denominator, or excuse me, the numerator.

02:12 So we have these two constants, the root 2 and pi, that we can bring outside of the integral.

02:17 So we have root 2 times pi times the integral from 3 up to 9 of root x minus 1, times the square root of 1 plus 1 over 8 times x minus 1.

02:34 Next thing that we're going to want to do here for doing the integration, let's make the substitution u equals x minus 1, and d x will be the same thing as d u.

02:46 So this will turn into root 2 pi as the integral from 3 up to 9 of the square root of u times the square root of 1 plus 1 over 8 u, du.

03:02 We can then simplify the radicals down a little bit, rewriting this as, or rewriting that second one, as u plus 1 over 8 divided by u, which, when we multiply by that square root of u outside, we can turn this into just the square root of u plus 1 over 8.

03:25 So, having that, we can do another substitution here.

03:32 Say, s equals, well, actually, s is already used, we'll do a little script s...

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