# Surface area of a curve around the x-axis

• compliant
In summary, we are asked to find the surface area traced out when the curve 8y^2 = x^2(1-x^2) is revolved around the x-axis. This can be solved by isolating y and using the formula for surface area of a curve rotated around an axis. The resulting integral can be simplified using the chain rule and the final answer can be obtained.
compliant

## Homework Statement

Find the surface area traced out when the curve 8y^2 = x^2(1-x^2) is revolved around the x-axis.

## The Attempt at a Solution

x-axis means y = 0
When y = 0, x = 0, -1 or 1.
Since this curve is "the infinity symbol", the curve has symmetry at x = 0.

Isolating y,
$$8y^2 = x^2(1-x^2)$$
$$y = \frac{\sqrt{x^2(1-x^2)}}{8}$$

Surface area = $$2\pi \int_{C} y dr$$
= $$4\pi \int^{1}_{0} y dr$$

$$dr = {\sqrt{1+ ({\frac{\sqrt{2x(1-2x^2)}}{8})^2}} {dx}$$
$$dr = {\frac{\sqrt{64+2x(1-2x^2)}}{8}} {dx}$$

Surface area = $${\frac{\pi}{8}} \int^{1}_{0} {\sqrt{x^2(1-x^2)(64+2x(1-2x^2))}{dx}$$And...I have no idea how to proceed with the integral.

Last edited:
Hi compliant!

(have a square-root: √ and an integral: ∫ and try using the X2 tag just above the Reply box )
compliant said:
Find the surface area traced out when the curve 8y^2 = x^2(1-x^2) is revolved around the x-axis.
Surface area = $$2\pi \int_{C} y dr$$ …

Nooo … using vertical slices, surface area = ∫2πydx times a factor to take account of the slope, which is … ?

The factor is $$\sqrt{1+(dy/dx)^2}$$, which I'm sure I accounted for above.

$$dr = {\sqrt{1+ ({\frac{\sqrt{2x(1-2x^2)}}{8})^2}} {dx}$$

It might not be "dr", but I'm 99% sure that's the calculation of the factor.

oh i see … yes that's right … i didn't recognise it from your:
compliant said:
$$dr = {\sqrt{1+ ({\frac{\sqrt{2x(1-2x^2)}}{8})^2}} {dx}$$

start again … if y = (1/√8)x√(1 - x2), then dy/dx = … ?

tiny-tim said:
oh i see … yes that's right … i didn't recognise it from your:

start again … if y = (1/√8)x√(1 - x2), then dy/dx = … ?

Ah, I must've forgotten to chain rule it.

y = (1/√8)√(x2 - x4)
dy/dx = (1/2√8)(x2 - x4)-1/2(2x - 4x3)
dy/dx = (1/√8)(x2 - x4)-1/2(x - 2x3)

compliant said:
dy/dx = (1/√8)(x2 - x4)-1/2(x - 2x3)

yes, that looks right … (you had a factor 2 earlier) …

now square it and add the 1, and I think it'll be a perfect square on the top

Ok, got the answer. Thanks.

## 1. What is the definition of "surface area of a curve around the x-axis"?

The surface area of a curve around the x-axis refers to the total area of the 3-dimensional shape that results when a 2-dimensional curve is rotated around the x-axis. This concept is commonly used in calculus to calculate the surface area of a solid of revolution.

## 2. How is the surface area of a curve around the x-axis calculated?

The surface area of a curve around the x-axis is calculated using a mathematical formula, known as the "surface area of revolution" formula. This formula involves integrating the square of the curve's derivative with respect to x, and then multiplying by 2π. The resulting value represents the total surface area of the shape.

## 3. Can the surface area of a curve around the x-axis be negative?

No, the surface area of a curve around the x-axis cannot be negative. Since surface area is a measure of the total area of a shape, it can never have a negative value.

## 4. What factors can affect the surface area of a curve around the x-axis?

The surface area of a curve around the x-axis can be affected by the shape of the curve, the interval of integration, and the axis of rotation. A larger interval of integration and a larger axis of rotation can result in a larger surface area, while a more complex curve may have a larger surface area compared to a simpler curve.

## 5. How is the concept of "surface area of a curve around the x-axis" used in real life?

The concept of surface area of a curve around the x-axis is used in several real-life applications, such as in engineering, architecture, and physics. For example, it can be used to calculate the surface area of a rotating object, such as a Ferris wheel or a propeller, or to determine the surface area of a curved roof or dome-shaped structure.

• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
458
• Calculus and Beyond Homework Help
Replies
2
Views
533
• Calculus and Beyond Homework Help
Replies
10
Views
426
• Calculus and Beyond Homework Help
Replies
14
Views
222
• Calculus and Beyond Homework Help
Replies
6
Views
843
• Calculus and Beyond Homework Help
Replies
4
Views
683
• Calculus and Beyond Homework Help
Replies
20
Views
449
• Calculus and Beyond Homework Help
Replies
5
Views
755
• Calculus and Beyond Homework Help
Replies
8
Views
869