Volume of Revolved Area: Washer Method Problem 1 | x=y^2, x=4, y=2

[elitespart]

1. Find the volume generated by revolving the area bounded by x=y$$^{2}$$ and x=4 about the line y=2.


Is it $$\pi\int(4-(2-\sqrt{x})^{2}dx$$ from 0 to 4?

 

[Dick]

No. The inner radius is (2-sqrt(x)). What's the outer radius? It doesn't lie on the x-axis.

 

[elitespart]

Yes, if you fix the unmatched parenthesis which makes it look a little confusing. The outer radius of the washer is 2 and the inner radius is (2-sqrt(x)).

oh whoops forgot to add a bracket at the end. Thanks for your help.

 

[Dick]

oh whoops forgot to add a bracket at the end. Thanks for your help.

Hold it. I gave it a second thought and edited my answer. The area enclosed by x=4 and y^2=x has pieces both about and below the x-axis.

 

[elitespart]

So it's 4 instead of 2?

 

[Dick]

No, y^2=x gives y=sqrt(x) or y=-sqrt(x). Those are the boundaries of the region. Your original answer would be ok if they had specified y=0 as a boundary. But they didn't.

 

[elitespart]

No, y^2=x gives y=sqrt(x) or y=-sqrt(x). Those are the boundaries of the region. Your original answer would be ok if they had specified y=0 as a boundary. But they didn't.

oh okay. So would y=2 factor into the outer radius too? Meaning r = [2 - (-sqrt(x))]?

 

[Dick]

oh okay. So would y=2 factor into the outer radius too? Meaning r = [2 - (-sqrt(x))]?

Exactly. The outer radius is 2+sqrt(x).

 

[elitespart]

awesome. Thanks again.