<p>I am complete stuck on trying to solve these problems.
1.Find the volume of the solid generated by the revolving the region bounded by the graphs of the equations about the indicated lines.
y=2x2, y=0, x=2
a) the y-axis b) the x-axis
c) the line y=8 d) the line x=2</p>
<p>2.find the volume of the solid generated by the revolving the region bounded by the graphs of the equations about the line y=4
y=x, y=3, x=0</p>
<ol>
<li>Find the volume of the solid generated by the revolving the region bounded by the graphs of the equations about the line x=6
y=x,y=0,y=4,x=6</li>
</ol>
<p>Could someone break it down and explain how to do these I'm kind of lost.</p>
<p>Are you familiar with the Washer and Disc Methods of Integration? If you are, these are situations in which you would apply them. I can detail them further if you have not heard of them.</p>
<p>I haven’t, could you explain it?</p>
<p>For the b) , you need to find where the bounds are. One is given at x=2, the other should be where y=2x^2 and y=0 intersect. This point is x=0. So the first regions bounds are 0,2.</p>
<p>Since we are revolving this, we are forming disks. The area of a circle is pi*r^2, therefore when we set up this integral we will square the radius (2x^2) and use the bounds 0,2 and then multiply the end product by pi.</p>
<p>pi<em>integral:(2x^2)^2 from 0,2
pi</em>integral:4x^4 from 0,2
pi<em>(4</em>2^4-0)=pi<em>(4</em>16)=64*pi</p>
<p>Someone correct me if I am wrong.</p>
<p>What about the other parts?</p>
<p>1a is 16pi
pi* integral from 0 to 8(2^2 - sqrt((y/2))^2)
is
16 pi</p>
<p>1b is NOT 64 pi(as explained above - he forgot to integrate)
it is (128/5)*pi</p>
<p>1c is (512/5)<em>pi
*it might be (1024/15)</em>pi
I’m getting the first, calculator is getting the second.</p>
<p>Could you break down how you solved that?</p>
<p>1a is 16pi
pi* integral from 0 to 8(2^2 - sqrt((y/2))^2)
is
16 pi</p>
<p>1b is NOT 64 pi(as explained above - he forgot to integrate)
it is (128/5)*pi</p>
<p>1c is (512/5)<em>pi
*it might be (1024/15)</em>pi
I’m getting the first, calculator is getting the second.</p>
<p>1d is (16/3)pi
calculator is getting (40/3)*pi</p>
<p>If you have the answers can you tell me if these 4 are right? I’ll explain them once they are confirmed.</p>
<p>Yep, sure did forget to integrate. Sorry. I would follow this link: [Khan</a> Academy](<a href=“Khan Academy | Free Online Courses, Lessons & Practice”>Khan Academy | Free Online Courses, Lessons & Practice)</p>
<p>Then go to solid of revolution videos.</p>
<p>4x^5/5
128/5*pi
yep lol.</p>
<p>Disc Method: for use when the side of the function against the line of revolution touches it for the entire interval on which you are taking the integral.</p>
<p>Think about the area of a circle (pi<em>r^2) but applied to the function, which is being rotated circularly. Thus, the integral would be pi</em>integral[f(x)]^2dx from (a,b)</p>
<p>Washer Method: for use when there is a “gap” between the area between the two functions and the line of revolution.</p>
<p>You are essentially finding the area under the outer curve all the way to the line of revolution and subtracting the area between the inner curve and the line of revolution, so you would have pi*integral[R^2-r^2]dx fro (a,b).</p>
<p>These methods and their variations (such as for lines of revolution not on the axes) are very difficult to explain without drawing pictures, using the correct symbols, etc. But, if you ask your teacher or somebody at your school who could help, these concepts are really not that difficult to learn. Good luck.</p>
<p>The Khan Academy site that was posted above is an amazing site.</p>
<p>Thanks, I’m going to watch. But could someone still help still need help.</p>