Thread: Volume of solid revolution

This is a little different from the other four I have done. The problem states:
Y=7x^2, x=1, y=0
about the x-axis

Some of the other problems have y=x^2 and y=3x and they tell you it is rotated around the line x=3. I know with this one we put x^2=3x then solve for x and get o,3 then put them back in to get the y's so we have (0,0) and (3,9) and now we know our limit is from 0 to 9 and then on our picture on the graph we know this volume is rotated around the line x=3. So we then integrate and then take the limit to get the area. I guess that this probelm looks a bit different but I still need to follow the same procedure. I tried to solve for x and had x= sqrt(y/7), x=1. I looked in the book for the odd number questions, so I could see if there were any like this worked out so I could get an idea what to do but as it usually goes, there are no examples like this. The next three are like this one, so any help would be appreciated.
Thank You,
Keith

Originally Posted by keith

This is a little different from the other four I have done. The problem states:
Y=7x^2, x=1, y=0
about the x-axis

Some of the other problems have y=x^2 and y=3x and they tell you it is rotated around the line x=3. I know with this one we put x^2=3x then solve for x and get o,3 then put them back in to get the y's so we have (0,0) and (3,9) and now we know our limit is from 0 to 9 and then on our picture on the graph we know this volume is rotated around the line x=3. So we then integrate and then take the limit to get the area. I guess that this probelm looks a bit different but I still need to follow the same procedure. I tried to solve for x and had x= sqrt(y/7), x=1. I looked in the book for the odd number questions, so I could see if there were any like this worked out so I could get an idea what to do but as it usually goes, there are no examples like this. The next three are like this one, so any help would be appreciated.
Thank You,
Keith

The intersection point of x = 1 and is (1, 7). So using the washer method:


-Dan

So, I solve for y=7x^2 and get y=7 but where did you get 2pi from? I would have just used pi. It is because it is a different shape?
Thankx,
Keith

Originally Posted by keith

So, I solve for y=7x^2 and get y=7 but where did you get 2pi from? I would have just used pi. It is because it is a different shape?
Thankx,
Keith

Sorry, I goofed! (I'll fix it in the original post.) The y coordinate is equal to the radius of the disc, so it should have been simply as you say. However I should have squared the y coordinate since we need the area of the washer.

-Dan

I think I got it. So after I take the integral I have pi*(49/5)x^5 and then I take the limit from 1 to 0 and get 9.8pi.
Thankx
Keith