Surface Area of y=sin^2(x)+x^2 from 0 to 1 about x axis

In summary, the formula for calculating the surface area of a curve about the x axis is S = ∫<sub>0</sub><sup>1</sup> 2πy√(1+(dy/dx)^2)dx. Each term in the formula represents a specific aspect of the surface area calculation, such as the integral representing the sum of infinitely small rectangles. The derivative of y=sin^2(x)+x^2 is dy/dx = 2sin(x)cos(x) + 2x. Integrating from 0 to 1 represents finding the surface area over a specific interval, and this formula has many real-life applications in fields like engineering, architecture, and physics.
  • #1
Potatochip911
318
3

Homework Statement


Set up a definite integral for the surface area generated by rotating the curve ##y= \sin^2x+x^2## from ##x=0## to ##x=1## about the a-axis.

Homework Equations


Surface Area about x axis=##2 \pi y \cdot ds ##

The Attempt at a Solution


I found ##\dfrac{dy}{dx} = 2 \sin x\cos x + 2x ## and ##ds= \sqrt{1+(\dfrac{dy}{dx})^2} \cdot dx ##
Therefore I got the surface area being equal to $$ SA= 2 \pi \int_{0}^{1} (\sin^2x+x^2) \sqrt{1+(2\sin x \cos x +2x)^2} \cdot dx $$ My professor however is saying the surface area is $$ SA= 2 \pi \int_{0}^{1} (\sin^2 x +x^2) \cdot dx $$ That doesn't make any sense to me since he's clearly not using the surface area formula. I was just wondering whether or not I did this correctly.
 
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  • #2
Potatochip911 said:

Homework Statement


Set up a definite integral for the surface area generated by rotating the curve ##y= \sin^2x+x^2## from ##x=0## to ##x=1## about the a-axis.

Homework Equations


Surface Area about x axis=##2 \pi y \cdot ds ##

The Attempt at a Solution


I found ##\dfrac{dy}{dx} = 2 \sin x\cos x + 2x ## and ##ds= \sqrt{1+(\dfrac{dy}{dx})^2} \cdot dx ##
Therefore I got the surface area being equal to $$ SA= 2 \pi \int_{0}^{1} (\sin^2x+x^2) \sqrt{1+(2\sin x \cos x +2x)^2} \cdot dx $$ My professor however is saying the surface area is $$ SA= 2 \pi \int_{0}^{1} (\sin^2 x +x^2) \cdot dx $$ That doesn't make any sense to me since he's clearly not using the surface area formula. I was just wondering whether or not I did this correctly.

I think your professor is having a bad day. You definitely need the derivative factor.
 

What is the formula for calculating the surface area of y=sin^2(x)+x^2 from 0 to 1 about the x axis?

The formula for calculating the surface area of y=sin^2(x)+x^2 from 0 to 1 about the x axis is given by:

S = ∫01 2πy√(1+(dy/dx)^2)dx

What does each term in the formula for surface area represent?

The integral (∫) represents the sum of infinitely small rectangles, 2π represents the circumference of a circle, y represents the function value at a specific point, √(1+(dy/dx)^2) represents the arc length of the curve, and dx represents the infinitesimal width of each rectangle.

How do you find the derivative (dy/dx) of y=sin^2(x)+x^2?

The derivative of y=sin^2(x)+x^2 is given by:

dy/dx = 2sin(x)cos(x) + 2x

What is the significance of integrating from 0 to 1 in the formula for surface area?

Integrating from 0 to 1 in the formula for surface area represents finding the surface area of the curve over a specific interval (in this case, from x=0 to x=1). This interval can be adjusted to find the surface area over a different range of x values.

What are some real-life applications of calculating surface area about the x axis?

Calculating surface area about the x axis has many real-life applications, such as in engineering, architecture, and physics. It can be used to determine the surface area of objects with curved surfaces, such as 3D-printed objects or buildings with curved roofs. It can also be used in the calculation of the surface area of body parts, such as the surface area of the brain or lungs. In physics, surface area calculations are used in the study of heat transfer and fluid dynamics.

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