1. ## Volume of a solid; please check

Okay, so, I think I am getting the hang of these problems. Would anyone care to check? All we are required to do for the homework is set up the integral.

$\displaystyle x=\sqrt{y}, x=0, y=1,$ ; about the x-axis

I set up my graph. It is the graph of a parabola with line y=1 going horizontally through it, and x=0. Seeing this, and the fact that it was solved in terms of y, I set up the integral in terms of a R and L volume. Integral is 0 to 1.

I got:

$\displaystyle 2\pi$ $\displaystyle \int |x-1| (\sqrt{y})dy$

Am I correct?

2. Originally Posted by Marconis
Okay, so, I think I am getting the hang of these problems. Would anyone care to check? All we are required to do for the homework is set up the integral.

$\displaystyle x=\sqrt{y}, x=0, y=1,$ ; about the x-axis

I set up my graph. It is the graph of a parabola with line y=1 going horizontally through it, and x=0. Seeing this, and the fact that it was solved in terms of y, I set up the integral in terms of a R and L volume. Integral is 0 to 1.

I got:

$\displaystyle 2\pi$ $\displaystyle \int |x-1| (\sqrt{y})dy$

Am I correct?
I'm afraid not ... when integrating w/r to a single variable, the integrand needs to be solely a function of that variable.

here are two set ups rotating the described region about the x-axis ...

using washers w/r to x ...

$\displaystyle \displaystyle V = \pi \int_0^1 1^2 - (x^2)^2 \, dx$

using cylindrical shells w/r to y ...

$\displaystyle \displaystyle V = 2\pi \int_0^1 y \cdot \sqrt{y} \, dy$

3. Shoot, three things I accidentally typed in:

For one, in my absolute value, I meant to put y. Secondly, I also accidentally put 1 for k because of the y=1. Really, though, k=0. Thirdly, it should be |0-y|, not |y-0| If I didn't make three (critical) mistakes, then I would have gotten your second answer.

Thank you!