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

## 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.

Dick
Homework Helper

## 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.