Find the volume of the solid revolved around a region

  • Thread starter Thread starter yesiammanu
  • Start date Start date
  • Tags Tags
    Solid Volume
yesiammanu
Messages
27
Reaction score
0

Homework Statement


Find the volume of the solid generated by revolving the region bounded by the graphs of y2=4x, the line y=x, about
A) x=4
B) y=4


So first I start out by graphing it
FwU0Q.png


The intercepts are at 0,0 and 4,4

I use the washers method since there is a gap in between the line and the rotated solid, making a space in the middle of the solid

The washers method says V= ∏∫ ([R(x)]2 - [r(x)2) where R(x) is the largest(outer) area, and r(x) is the smallest(inside) area. This is for rotating about the x-axis but can be used to rotate around the y axis. However, this isn't rotating about either of these axis; rather, it's rotating around x=4 which is what I am having trouble with.

So boundaries are 0 to 4, equation is ∏∫([y2/4]2 - y2)dy but this is wrong. How would I do it correctly? Is cylindrical shells a better method? If I was using cylindrical shells, would it be 2∏∫(2√x - x)(x) since I use ∫ 2∏(shell height)(shell radius) Also I'm a bit confused on how I would do part B

Any help would be great, thanks.
 
Last edited:
Physics news on Phys.org
For part B, also using the cylindrical shells method, would the proper integral be 2∏∫(y - y2/4)(y)
 
yesiammanu said:

Homework Statement


Find the volume of the solid generated by revolving the region bounded by the graphs of y2=4x, the line y=x, about
A) x=4
B) y=4


So first I start out by graphing it
FwU0Q.png


The intercepts are at 0,0 and 4,4

I use the washers method since there is a gap in between the line and the rotated solid, making a space in the middle of the solid

The washers method says V= ∏∫ ([R(x)]2 - [r(x)2) where R(x) is the largest(outer) area, and r(x) is the smallest(inside) area. This is for rotating about the x-axis but can be used to rotate around the y axis. However, this isn't rotating about either of these axis; rather, it's rotating around x=4 which is what I am having trouble with.

So boundaries are 0 to 4, equation is ∏∫([y2/4]2 - y2)dy but this is wrong. How would I do it correctly? Is cylindrical shells a better method? If I was using cylindrical shells, would it be 2∏∫(2√x - x)(x) since I use ∫ 2∏(shell height)(shell radius) Also I'm a bit confused on how I would do part B

Any help would be great, thanks.

Nice picture. For the first one, if you are rotating around x=4, then wouldn't the inner radius be 4-y and the outer radius be 4-y^2/4? The radii should be distances to the axis of rotation.
 
Ah I see, that makes sense, thank you. Is what I posted above (2∏∫(2√x - x)(x) for x=4, 2∏∫(y - y2/4)(y) for y=4) correct for cylindrical shell method?
 
yesiammanu said:
Ah I see, that makes sense, thank you. Is what I posted above (2∏∫(2√x - x)(x) for x=4, 2∏∫(y - y2/4)(y) for y=4) correct for cylindrical shell method?

Same problem. I think you have the lengths of the shells right, but you don't have the radius right. You aren't rotating around 0.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top