Surface area of a rotated curve

• glid02
In summary, the conversation involved finding the area of a surface obtained by rotating a curve around the x-axis. The suggested method was to parametrize the curve using rcos(theta) as x and rsin(theta) as y, but it was pointed out that this would result in a circle. The correct parametric equations were given, and the process for finding the area was explained, involving finding partial derivatives, taking their cross product, and integrating. An alternative, simpler method involving the surface area of a cylinder was also suggested.
glid02
Here's the question:
Find the area of the surface obtained by rotating the curve

from x=0 to x=9 about the x-axis.

I'm supposed to parametrize the curve, using rcos(theta) as x and rsin(theta) as y, at least I think I am.

That would give f(x,y) = 3rcos^3(theta),rsin(theta)

Then find the partial derivatives with respect to r and theta and find their cross product. Then find the magnitude of the cross product and integrate with limits int[0-2pi] int[0-9].

Is this right? I can't find any information on the internet to do it this way and the book isn't much help either.

Thanks.

Last edited by a moderator:
glid02 said:
Here's the question:
Find the area of the surface obtained by rotating the curve

from x=0 to x=9 about the x-axis.

I'm supposed to parametrize the curve, using rcos(theta) as x and rsin(theta) as y, at least I think I am.
No. That is a circle in the xy-plane. Your curve, in the xy-plane, is y= 3x3, not at all a circle. When you "rotate" around the x-axis, every point, with coordinates (x0, y) in the xy-plane, traces out a circle of points in the (x0, y, z) plane.
Since the radius of the circle is y= 3x^3, that would have parametric equations $x= x_0, y= 3x_0^3cos(\theta ), z= 3x_0^3 sin(\theta)$.

That would give f(x,y) = 3rcos^3(theta),rsin(theta)

Then find the partial derivatives with respect to r and theta and find their cross product. Then find the magnitude of the cross product and integrate with limits int[0-2pi] int[0-9].

Is this right? I can't find any information on the internet to do it this way and the book isn't much help either.

Thanks.
Using the parametric equations I gave, with x0 and $\theta$ as parameters, that should work. However, it looks awfully complicated to me. The area of the curved surface of a cylinder with radius y and height ds is $2\pi y ds$. For y= 3x3,
$$ds= \sqrt{1+ (9x^2)^2}dx= \sqrt{1+ 81x^4}dx$$.
The integral for surface area would be given by
$$2\pi \int_0^9(3x^3)\sqrt{1+81x^4}dx$$

Last edited by a moderator:

What is the definition of surface area of a rotated curve?

The surface area of a rotated curve is the amount of area that is covered by the curve when it is rotated around a specific axis. This is often calculated in terms of square units, such as square inches or square meters.

How is the surface area of a rotated curve calculated?

To calculate the surface area of a rotated curve, one must use a mathematical formula that takes into account the shape and dimensions of the curve, as well as the axis of rotation. This formula is typically derived using integral calculus.

What is the significance of the surface area of a rotated curve in real life?

The surface area of a rotated curve has many practical applications, such as determining the amount of material needed to coat a curved object, or calculating the surface area of a 3-dimensional shape in engineering and manufacturing processes.

How does the surface area of a rotated curve relate to the concept of volume?

The surface area of a rotated curve is closely related to the concept of volume, as both involve calculating the amount of space occupied by a 3-dimensional object. However, while surface area measures the outer covering of the object, volume measures the total amount of space inside the object.

Can the surface area of a rotated curve be negative?

No, the surface area of a rotated curve cannot be negative. Since surface area is a measure of the amount of space occupied by an object, it cannot have a negative value. It is always a positive value, or zero if the curve has no surface area (such as a straight line).

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