# Homework Help: Volume computation (from cross-sectional area)

1. Jul 5, 2010

### BrownianMan

The base of each solid below is the region in the xy-plane bounded by the x-axis, the graph of $$$y=\sqrt{x}$$$ and the line x = 3. Find the volume of each solid.

a) Each cross section is perpendicular to the y-axis is a square with one side in the xy-plane.
b) Each cross section is perpendicular to the y-axis is an isosceles right triangles with hypotenuse in the xy-plane.

This is what I have:

a)

$$$V=\int_{0}^{\sqrt{3}}y^4 \ dy=\frac{1}{5}y^5\biggr|_{0}^{\sqrt{3}}=\frac{9\sqrt{3}}{5}$$$

b)

$$hyp=y^2$$

$$$A=\frac{1}{2}\left ( \frac{y^2}{\sqrt{2}} \right )^2=\frac{y^4}{4}$$$

$$$V=\frac{1}{4}\int_{0}^{\sqrt{3}}y^4 \ dy=\frac{1}{20}y^5\biggr|_{0}^{\sqrt{3}}=\frac{9\sqrt{3}}{20}$$$

However, the answer in the back of the book is $$$\frac{24\sqrt{3}}{5}$$$ for a) and $$$\frac{6\sqrt{3}}{5}$$$ for b). What am I doing wrong?

Last edited: Jul 5, 2010
2. Jul 5, 2010

### The Chaz

a. the base of the square isn't (y^2); it's (3-y^2).