Solids of revolution volume

  • Thread starter PhizKid
  • Start date
  • #1
475
1

Homework Statement


Volume of the region bounded by y = x^2 and x = y^2 about y = 1


Homework Equations


[itex]\pi r^2[/itex]


The Attempt at a Solution


So the functions look something like this:

ICX4Rud.png


I decided to use method of washers with respect to x.

The radius if the center is at y = 1 of the washers is going to be [itex]\sqrt{x} - x^2[/itex] and the inside space is going to have a radius of [itex]1 - \sqrt{x}[/itex]. So we want to subtract these 2 volumes:

[itex]\pi\int_{0}^{1}(\sqrt{x} - x^2)^2 dx - \pi\int_{0}^{1}(1 - \sqrt{x})^2 dx[/itex]

So I get:

[itex]\pi\int_{0}^{1}(\sqrt{x} - x^2)^2 dx - \pi\int_{0}^{1}(1 - \sqrt{x})^2 dx \\
\pi\int_{0}^{1}(x - 2x^{\frac{5}{2}} + x^4) - (1 - 2x^\frac{1}{2} + x) dx \\
\pi\int_{0}^{1}(-2x^{\frac{5}{2}} + x^4 - 1 + 2x^\frac{1}{2}) dx \\
\pi[(\frac{-4}{7}x^{\frac{7}{2}} + \frac{x^5}{5} - x + \frac{4}{3}x^\frac{3}{2})]_{0}^{1} \\
\pi(-\frac{4}{7} + \frac{1}{5} - 1 + \frac{4}{3}) \\
\pi\frac{-4}{105}[/itex]

Not sure how I get a negative number. What did I do wrong?
 

Answers and Replies

  • #2
eumyang
Homework Helper
1,347
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I decided to use method of washers with respect to x.

The radius if the center is at y = 1 of the washers is going to be √x - x2 and the inside space is going to have a radius of [itex]1 - \sqrt{x}[/itex].
I don't think the outer radius is correct. We want the distance from the curve y = x2 to the line y = 1. Shouldn't it be 1 - x2?
 
  • #3
475
1
Oh, I was using the wrong radius...thanks, I see it now
 

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