- #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:

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?