# Calculating Surface Area around a Vertical Line: Tips and Tricks"

• yeuVi
In summary, the function has a surface area that is described by the integral of the radical term and the distance between the line segment and the axis of rotation. If you substitute y with (y+3), you need to change the circumference of the circle around the axis for that value of y.
yeuVi

How do you find a surface area of a function around a vertical line?

For example: surface of $$\ y=x^3$$ between $$\ 0<= x <= 3$$, rotating around $$\ x=4$$?

I tried the formula for finding surface area, but I confuse with that vertical line x=4... what should I do?

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How would you set up the integral if it asked you to rotate it around the y-axis and what change can you make to account for this difference?

If around the y-axis, then the integral should be

$$\int 2\pi*y^{1/3}* \sqrt{1+ {1/9}*y^{-4/3}}*dy$$

right? From then, what change should I make to adjust with that x=3 vertical line?

Well, I just need to know how to adjust that integral with the vertical line at x=3... the formula is easy, but the interval of integral should be ...?

yeuVi said:
If around the y-axis, then the integral should be

$$\int 2\pi*y^{1/3}* \sqrt{1+ {1/9}*y^{-4/3}}*dy$$

right? From then, what change should I make to adjust with that x=3 vertical line?

Your integral for around the axis x = 0 contains the radical term that represents the line segment along the curve from y to y + dy. It also contains a term that represents the distance of that line segment from the axis of rotation related to computing the circumference of the circle around the axis. If you move to a new axis, the thing that changes is the circumference of the circle around the axis of rotation. How far are you from the new axis of rotation for each value of y?

Im sorry... You lost me... I guess what u try to say is that I substituted y with (y+3) in the integral formula?

yeuVi said:
Im sorry... You lost me... I guess what u try to say is that I substituted y with (y+3) in the integral formula?

The integral you wrote for rotation around x = 0 was

$$\int 2\pi*y^{1/3}* \sqrt{1+ {1/9}*y^{-4/3}}*dy$$

The part of the integrand that is

$$2\pi*y^{1/3$$

is the circumference of a circle around the x = 0 axis of rotation corresponding to any particular value of y = x^3. The radical term times dy is the length of the curve y = x^3 between a particular value of y and y + dy. Isn't that how you found the integrand?

What you must change is not the length of the curve between y and y + dy. You must change the circumference of the circle around the axis for that value of y. The new radius of the circle would be (4 - x) expressed in terms of y. Do you see why? Draw the curve and the axis of rotation.

Thank you very much... now I got the problem...^_^

## 1. What is surface area?

Surface area is the total area of all the surfaces of a three-dimensional object. It is measured in square units, such as square meters or square inches.

## 2. How do you calculate surface area?

The formula for calculating surface area depends on the shape of the object. For example, the surface area of a cube is found by multiplying the length of one side by six, while the surface area of a cylinder is calculated by multiplying the circumference of the base by the height and adding the area of the top and bottom circles.

## 3. Why is surface area important?

Surface area is important because it helps us understand the amount of material needed to cover or coat an object. It also plays a role in many scientific and engineering applications, such as heat transfer and chemical reactions.

## 4. What is the difference between surface area and volume?

Surface area is the measure of the total area of all the surfaces of an object, while volume is the measure of the space inside an object. In other words, surface area is the measurement of the outside of an object, while volume is the measurement of the inside.

## 5. How does surface area affect the properties of an object?

The surface area of an object can affect its properties in several ways. For example, a larger surface area can increase the rate of heat transfer, while a smaller surface area can decrease it. In chemical reactions, a larger surface area can increase the surface area for reactants to come in contact, leading to a faster reaction rate. In terms of strength, a larger surface area can distribute weight and force more evenly, making an object stronger.

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