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.)

Question

owlfred · Answer

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

watchmath · Answer

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

anonymous · Answer

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

watchmath · Answer

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

watchmath · Answer

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

watchmath · Answer

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

watchmath · Answer

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