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

step 1 Okay, what we want to do is this is kind of a longer problem. A, B, and C. B and C are going to

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

Key Concepts

Prolate Spheroid

A prolate spheroid is a type of ellipsoid generated by revolving an ellipse around its major axis. This shape is elongated compared to a sphere, having a longer axis along the direction of rotation. Important formulas related to the prolate spheroid include those for volume and surface area, which involve the semi-major and semi-minor axes of the generating ellipse and often incorporate elliptic integrals in the surface area calculation.

Oblate Spheroid

An oblate spheroid is formed by revolving an ellipse about its minor axis, resulting in a shape that is flattened along the axis of rotation. Like the prolate spheroid, its volume can be computed using formulas involving its semi-major and semi-minor axes, but the surface area calculation often requires different techniques, such as approximations or closed-form expressions involving logarithmic or trigonometric functions, due to the complexity introduced by the flattening.

Ellipse

An ellipse is a conic section defined as the set of points for which the sum of the distances to two fixed points (the foci) is constant. In its standard form, an ellipse is typically written as x²/a² + y²/b² = 1, where a and b represent the lengths of the semi-major and semi-minor axes, respectively. Its area can be calculated using the formula A = ?ab, which is a foundational result in conic sections and geometry.

Solid of Revolution

A solid of revolution is a three-dimensional object obtained by rotating a two-dimensional region about a specified axis. This process is significant in calculus and engineering because it allows for the computation of volumes and surface areas using methods such as the disk, washer, or shell methods. When a region bounded by a curve is rotated, the resulting object often has symmetrical properties that simplify the calculations.

Transcript

00:01 Okay, what we want to do is this is kind of a longer problem.

00:06 A, b, and c.

00:08 B and c are going to be the longest parts of it.

00:11 But the first thing we want to do for part a is we want to find the area of a region bounded by an ellipse.

00:27 And the equation for the ellipse is x squared over 16.

00:32 Plus y squared over nine equal to one.

00:37 And so what the ellipse quickly looks like is we have 4 and negative 4, and then we have 3 and negative 3.

00:50 And so the ellipse is just this guy right here, where the major axis is the x -axis and the minor axis is the y -axis.

00:59 And so what we want to do is basically we notice that these are symmetrical.

01:05 There's four symmetrical or equal pieces.

01:09 And so if i can find the area of this piece right here, then i can multiply it by four to find the entire area that is bounded by the ellipse.

01:19 And so that's what we're, that's our goal.

01:22 And so the first thing i need to recognize is i've got to get y by itself.

01:27 So i have y squared is equal to nine times one minus x squared over 16.

01:37 I'm going to go ahead and factor out a 116th.

01:42 So i have 9 over 16 times 16 minus x squared.

01:47 And then i'm going to take the square root of both sides.

01:50 So y is equal to plus and minus three -fourths times the square root of 16 minus x squared.

01:57 Squared.

01:58 Okay.

01:59 And so since i'm just working in this region right here, my area is just going to be the positive of the square root.

02:08 And so this is going to be equal to four times zero, the integral from zero to four, because that's my x value goes from zero to four, of the positive three force of the square root of 16 minus x squared and we're integrating with respect to x and so this becomes equal to three times the integral from zero to four of the square root of 16 minus x squared with respect to x and so it should be able to start recognizing the integral of the square root of a squared minus u squared or a squared minus x squared and so that is going to be equal to 3 times 1 half of x times the square root of 16 minus x squared plus 16 times the arc sign or inverse sign of x over 4 and we're going to evaluate it from zero at the upper limit of 4 and the lower halves.

03:22 So at 4 and 0, this entire part goes to 0.

03:27 And then at 0, this part, the 16 arc sign of 0 is 0 as well.

03:34 So i only have to worry about 16 times the arc sign of 1, which is pi over 2.

03:42 And so this becomes 12 pi.

03:45 So there's my area of the ellipse.

03:48 Okay, and so now we want to part b to find the volume and surface area of the region when it is revolved about the x axis.

04:17 So basically i want to find the region, the volume, and the region created when we revolve that ellipse about the, oh, that should say, x -axis or the major axis.

04:41 And so what that is going to look like is we know this goes from 3 to negative 3, our major axis.

04:51 As the x -axis.

04:54 And so we're going to take this, whoops, and we're going to revolve it about the x -axis.

05:01 Okay, and so basically when we do that, we get more of a football -shaped item.

05:08 And so the volume is going to be equal to, and basically what i'm going to be doing is i am, we're going to use the shell method and so here's my representative rectangle and i'm only need to revolve the region from this upper region right here to get the entire football shape item and so i'm going to do two because i only want to go from zero to four and then i double it to get this region from here to here as well so it's from zero two times two pie and i'm going to go from zero to four okay, and then of course it is your function squared because it's pi r squared, h and h is, or the width, which is dx.

06:03 And so this is going to be 9 over 16 because i'm squaring 3 fourth times 16 minus x squared dx.

06:14 And so my height is r or r squared is my function, and dx represents my width of my rectangle.

06:25 And so this is going to become equal to 9 pi over 8 times the integral from 0 to 4 of 16 minus x squared, integrating with respect to x.

06:43 So this is going to be 9 pi over 8 times 16x minus 1 3rd x cubed, and we're going to evaluate it at the upper limit of zero, i mean upper limit of four, subtract from it the lower limit of zero.

06:59 And so when i do that, i should get 48 pi.

07:05 And so really i'm only evaluating it at the upper limit.

07:10 Okay, now we got to talk about that surface area.

07:13 That surface area is going to be the issue.

07:15 And so we know that the surface.

07:21 Surface area is actually going to be the is given by the integral of 2 pi times the distance the rectangle is or the distance the function is from the axis of rotation.

07:52 So that's going to be my y value or my function.

07:57 And then it's times that function itself.

08:01 And so the surface area, or not times the function, but times the derivative.

08:07 So the surface area, the general form, is equal to 2 pi times the integral from a to b times y.

08:18 And this is the distance the graph is from axis of rotation.

08:34 And so since i'm rotating this way, it's going to be my y value, and then it's times the square root of 1 plus the derivative squared.

08:47 Okay, so basically i'm going for that format.

08:50 And so what i need to do, first of all, is to take the derivative, and square it.

08:56 And so we know that the function is y is equal to 3 4 times the square root of 16 minus x squared.

09:07 Okay.

09:07 And so i'm going to take the derivative.

09:09 So y prime is equal to negative 3x over 4 times the square root of 16 minus x squared.

09:18 And then i'm going to square that.

09:21 So i get 9x squared over 16.

09:25 Times 16 minus x squared.

09:29 Okay.

09:30 And so now my surface area is equal to, and i'm going to double it because i want to only integrate from zero to four.

09:41 So i still only want to integrate over this first part right here, this first part right here, and then double it to get the second part, and then i'm revolving it around this axis.

09:57 And so this is going to be equal to the y value, and the y value is 3 -4 times the square root of 16 minus x squared, times the square root of 1 plus 9x squared over 16 times, 16 minus x squared.

10:28 And we're integrating with respect to x.

10:31 Okay, so this is getting a little crazy.

10:35 And so i'm going to pull out this three -fourths, and so this becomes 3 pi times the integral from 0 to 4.

10:44 And then i have the square root of 16 minus x squared...

Please give Ace some feedback