anonymous 5 years ago write the definite integral that represents the area of the surface formed by revolving the graph f(x)=x^(1/2) on the interval [0,4] about the y-axis (Do not evaluate the integral.)

1. Owlfred

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2. watchmath

$\int_0^2 2\pi y^2\sqrt{1+(2y)^2}\,dy$

3. anonymous

because it is root x, wouldn't it turn in to x, not y squared?

4. watchmath

$$y=x^{12}$$ is equivalent to $$x=y^2$$

5. watchmath

I mean $$y={x^{1/2}}$$ is equivalent to $$x=y^2$$

6. watchmath

hold on.... I think I made a mistake

7. watchmath

So $$y=x^{1/2}$$ and $$dy/dx=(1/2)x^{-1/2}$$. So the surface area is $\int_0^42\pi x\sqrt{1+(1/4)x^{-1}}\, dx$

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