Find (a) the area of the region bounded by the ellipse, (b) the volume and surface area of the s

step 1 Okay, we are going to look at an ellipse. This ellipse is shown, and we are going to do several

Find (a) the area of the region bounded by the ellipse, (b)...

Key Concepts

Ellipse Geometry

An ellipse is a conic section characterized by two perpendicular radii: the semi?major axis and the semi?minor axis. Its standard equation provides these parameters, and the area of the ellipse can be directly computed using the formula ? multiplied by the product of these two axes. This concept underpins understanding the spatial extent of the region in question.

Region Area Calculation

Calculating the area of a bounded region, such as an ellipse, can be done using either geometric formulas or integration techniques. In the case of an ellipse, the area is found using the product of ? and the lengths of the semi-axes, reflecting an application of integral calculus in deriving known geometric results.

Solids of Revolution

Solids of revolution are generated when a plane region is revolved about an axis. Understanding this concept involves recognizing how rotating two-dimensional shapes yields three-dimensional objects, and it leads to the application of integration methods or known formulas for volume and surface area of the resulting solids.

Prolate and Oblate Spheroids

When an ellipse is revolved about one of its axes, the resulting three-dimensional shapes are spheroids. A prolate spheroid occurs when the region is rotated about the major axis, leading to a shape elongated along that axis, whereas an oblate spheroid results from rotation about the minor axis, yielding a shape flattened along the axis of revolution. These classifications are pivotal in applying the correct geometric formulas.

Volume and Surface Area Formulas for Spheroids

The computation of volume and surface area of spheroids involves specific formulas that depend on whether the spheroid is prolate or oblate. These formulas are derived using integration techniques or by modifying the formulas for spheres. Knowledge of these formulas is essential to accurately evaluate the solids resulting from the rotation of the ellipse.

Transcript

00:02 Okay, we are going to look at an ellipse.

00:05 This ellipse is shown, and we are going to do several things.

00:10 The first thing we're going to do is we're going to find the area of the region bounded by the ellipse.

00:16 So first of all, we just need to sketch this, just to remember how we're doing area in the future volume problems.

00:24 Remember, an ellipse is x squared over a squared, and a is how far the ellipse is going to be going from its center.

00:34 Now, before we go on, notice that there's no x minus a number inside the square or a y minus a number, so we are centered on zero, zero.

00:43 Okay, so four is two squared, so that means that our ellipse is going to go two units to the right of zero and two units to left, where our y will only go one up and one down.

00:55 So now that we've drawn a picture of the lips, now we can think about how to do area.

01:00 Area is all about top function minus bottom function, right? you're finding a height, and when you integrate it in terms of x, that is like an infinite number of rectangles.

01:14 So our top function here, we're going to need to solve for and the bottom function because the ellipse is not in function notation.

01:24 So let's go ahead and subtract that x -squarex.

01:27 Squared divided by four from both sides, and then go ahead and take the square root.

01:34 Now, when you take a square root, notice i have that plus or minus the square root.

01:39 So when this happens, the positive side is in quadrants one and two, and the negative side is in quadrants three and four.

01:47 So you can really see that our top function and our bottom function are really similar.

01:53 Just the bottom function has a negative in front of it.

01:56 So since we're integrating in terms of x, we're going to start at negative 2 and go to positive 2, and then we're writing that top function minus the negative for the bottom function.

02:12 Now we can clean that up a little bit and just have two of those because the minus the negative.

02:19 If this were to be integrated by hand, you would need to use some trig substitution.

02:25 However it does not look like that's necessary.

02:29 So our answer is 2 pi.

02:33 Our next question is going to ask us about volume and surface area.

02:40 And so we'll go ahead and just get some information from the last slide down.

02:46 Okay, so first we're going to revolve around the major axis and then later we're going to revolve around the minor axis.

02:53 So remember your major axis.

02:55 Is kind of your wider one.

02:58 And so here we're going to be revolving around the x -axis.

03:03 So volume is all about infinitely adding a whole bunch of circles together.

03:09 So that's all about pi r squared and they all have that thickness of dx.

03:14 So again we're going to be going from negative two to two.

03:18 Now our radius is just going to be that top function minus zero.

03:23 So we're not going to be having to deal with the negative side.

03:27 So we can just take that 1 minus x squared over 4, which was in a square root and square it, and now we no longer have that.

03:35 So this would be pretty simple to do just using power rule.

03:41 And we get 8 pi over 3.

03:44 Now surface area, and remember in this section, even though you're doing polar, you're having to remember some of your integration techniques from previous chapters.

03:53 So to remember that, surface area is a circumference, 2 pi r, so that's 2 pi times our function, but then it's also multiplied by a thickness.

04:04 But the thickness is what you would think of, like when you're doing pythagorean, it would be that hypotenuse.

04:11 So we have that extra piece in there.

04:14 So it looks like in order to do surface area, we are going to have to take the derivative so that we can place it into our problem.

04:27 So take the derivative.

04:29 I'm just going to clean up my original function here.

04:31 I'm going to find a common denominator and that was four...