Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If we've got two functions $Y=4$ and $y= \sqrt x$ which limits an area with the $Y$-axis. What is the rotation volume of that particular area.

My thoughts about the question is that i put the two equations against each other: $4 = \sqrt x$, $x = 16$.

So my limits are 0 - 16 which we use in the integral,

$2\pi \int x(4- \sqrt x) dx $

But the answer is wrong? How should i think?

share|improve this question
    
What answer did you get? Looks OK so far. –  Ross Millikan Jun 2 '11 at 12:44
    
I got 2948,9! is this right? –  Adam Jun 2 '11 at 13:20
    
No, I get a value less than 1/4 of that. –  Ross Millikan Jun 2 '11 at 13:31
    
ok, i corrected some of my calculations, finally i got 643,39...? is this right? –  aka Jun 2 '11 at 13:37
    
Are you expected to report the exact value or a decimal approximation? If decimal, it is fine. –  Ross Millikan Jun 2 '11 at 13:40

1 Answer 1

I take it that the region is rotating about the y-axis, right?

So effectively we have the curve $x=f(y)=y^2$ rotating in the interval $0\le y\le 4$. Therefore the volume is given by the integral

$$ \int_0^4\pi f(y)^2\,dy=\int_0^4\pi y^4\,dy=\frac{1024\pi}{5}. $$

Edit: Sorry, (new guy here) I should be extra careful in answering the question rather than just giving the answer to what might be a homework problem :-)

The way I think (and teach my students to think) about a problem like this is that whenever the curve is rotating about the $y$-axis instead of the more common $x$-axis, you should write the boundary curves in the form $x=f(y)$. In other words, try to find the inverse function. After all, the disk/washer method of computing volumes doesn't care, which axis is the axis of rotation. Or in other words, the shape and volume of the object does not change, if you switch the labels of the two axes!

share|improve this answer
    
I believe OP is using what's commonly known as "the washer method." –  Gerry Myerson Jun 2 '11 at 22:55
    
@Gerry: You are right! I had not seen that variant of the washer method earlier, but it is easy to justify the integral by dividing the object into washers that have an infinitesimally thin crust and a finite height as opposed to the (more common?) washers that have infinitesimal height. Wish my first year calculus students some luck next Spring ;-> –  Jyrki Lahtonen Jun 3 '11 at 13:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.