Draw the region bounded by the curves y=sqrt(x), x=4, y=0. Use the washer method to find the volume

Draw the region bounded by the curves y=sqrt(x), x=4, y=0. Use the washer method to find the volume

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Draw the region bounded by the curves Use the washer method to find the volume

We’re gonna draw the region bounded by the following equations and then we’re gonna take that region and rotate it about the Y-axis this time. Let’s take a look at this graph. The square root of X looks like this X equals four is a vertical line and Y equals zero is this X-axis here. So the region in question is this one and if we’re going to use the washer method to rotate it about the Y axis we’re going to slice this region up like this we’re gonna rotate it around the Y axis like this giving us a bunch of washers with the width of D Y. So let’s find that volume.
As always we get the volume of a region by adding up all the little volume elements. In other words we’re gonna add up little mini volumes of each one of these washers and as is always the case with the washer method the volume of each one of these disks is going to be PI times the outer radius squared minus the inner radius squared times in this case D Y. Now the outer radius of this washer is just going to be four no matter what. No matter where I slice this region to get a washer the outer radius is going to be given by the equation X equals four.

So in our volume equation we can plug in capital R equals four. Our inner radius is always going to be given by the equation Y equals square root of X. No matter where we slice this thing the inner radius is given by this function. Of course we’re integrating with respect to Y now so we need to solve for X in this equation to get a function of Y. We can solve for X by taking a square of both sides that’s going to give us X equals Y squared that inner radius of X equals Y squared can then be plugged in down here and now we can integrate that with respect to Y. Now what are our limits on Y. You’ll see down here in this region that the smallest value of Y is zero and the largest value of Y in this region happens right up here when we plug X equals four into the equation Y equals the square root of X. Well when we do plug X equals four into that equation we get Y equals tow. In other words this point here is the point four two so the largest value of Y for this region is two. All right now let’s integrate that and get an answer. Four squared root of sixteen, Y squared squared is Y to the fourth. We can integrate sixteen to get sixteen Y.
We can integrate Y to the fourth to get one-fifth Y to the fifth and we can plug in our limits which gives us something that isn’t too difficult to simplify. I”m getting thirty two minus thirty two fifths. Thirty two is the same thing as a hundred and sixty fifths and when we subtract thirty two fifths from that we get a hundred and twenty eight fifths PI. That is the volume of the region so that is our answer. As always I’ll zoom out for you so you can see the whole problem and as always I hope that that helped you.

About Post Author

Professor Cram

Professor Lawrence Cram is a Visiting Fellow at the Australian National University working in the Department of Applied Mathematics. His interests include astronomy, mathematics, engineering, computing, and physics. Due to his extensive expertise, professor Cram has worked as a Professor of Physics at the University of Sydney and as the Deputy Vice-Chancellor at ANU during 2004-2012. In 2013, he retired as a Master, University House and Graduate House. In January 2014, he was appointed as an acting Deputy Vice-Chancellor at Charles Darwin University. Professor Cram is also a Fellow at the Royal Astronomical Society and the Australian Institute of Physics.
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