Volume of a solid

[InterserveVB]

For some reason I can never seem to get the right answers. Can someone explain the following problems in detail?

Find the volume of a solid obtained rotating the region bound by x = y^2 and x = 4y about the y axis

Find the volume of the solid obtained by rotating the region bound by y = x^(1/5) and y = x about the line y = 1 (I have no clue how to take into consideration the y = 1)

 

[pka]

 

[Unco]

Find the volume of the solid obtained by rotating the region bound by y = x^(1/5) and y = x about the line y = 1 (I have no clue how to take into consideration the y = 1)

Draw a sketch (you don't need a computer) to see that you can translate both curves down 1 and rotate about the x-axis to obtain the same volume.

 

[galactus]

 

[InterserveVB]

Thx?

Thank you for your help with the second problem. I keep getting 2pi/7 is this correct? My book says the answer is 8pi/21 Is the book wrong? For the x^1/5. Also they used the 5 root of x. That is the same as x^1/5 correct?

 

[galactus]

\(\displaystyle {\pi}\int_{0}^{1}((x^{\frac{1}{5}})^{2}-x^{2})dx=\frac{8{\pi}}{21}\)

\(\displaystyle 2{\pi}\int_{0}^{1}y(y-y^{5})dy=\frac{8{\pi}}{21}\)

I believe the book translated down, as per Unco's suggested, and arrived at the answer.

 

[InterserveVB]

Ah ok, so 8pi / 21 is correct?

 

[galactus]

Since the two intersect at y=1 and the x-axis, the volume is the same regardless

of which they are revolved about. I suppose that is what they are getting at.

Yes, \(\displaystyle x^{\frac{1}{5}}\) is the fifth root of x.

Frankly, I would think the other integrals I had posted would be correct. Maybe

someone else has some ideas.

\(\displaystyle {\pi}\int_{0}^{1}((1-x)^{2}-(1-x^{\frac{1}{5}})^{2})dx=\frac{2{\pi}}{7}\)

\(\displaystyle 2{\pi}\int_{0}^{1}(1-y)(y-y^{5})dy=\frac{2{\pi}}{7}\)