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

#### Potatochip911

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Surface Area about x axis=$2 \pi y \cdot ds$

3. 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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#### Dick

Science Advisor
Homework Helper
1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Surface Area about x axis=$2 \pi y \cdot ds$

3. 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.

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