# Volume of cross sections using isosceles right triangle

1. May 28, 2012

### mikaloveskero

1. The problem statement, all variables and given/known data
The area is bounded by the equation y=x^2, x axis and the line x=3 and is perpendicular to the x axis

2. Relevant equations
A=1/2bh
Height=Base

3. The attempt at a solution
A=1/2B^2
V=1/2B^2 from 0 to 3
V=1/2 B^3/3 from 0 to 3
V=B^3/6 from 0 to 3
V=27/6

please tell me if im doing something wrong, on the right track or i'm correct

2. May 28, 2012

### HallsofIvy

Staff Emeritus
Are we to assume that the cross sections perpendicular to the x-axis are "isosceles right triangles"- with one leg perpendicular to the x-axis or the hypotenuse?

I think what you are trying to do is correct but your notation is very strange. You are given information about x and y- use them, not new variables! Assuming it is the leg that lies perpendicular to the x-axis then the other leg is just y= $x^2$. The area of such a triangle is $(1/2)(x^2)(x^2)= x^4/2$. The "thickness" of each such cross-section is $\Delta x$ so the volume is $(1/2)x^4\Delta x$. In the limit that becomes an integral.

3. May 28, 2012

### mikaloveskero

Its perpendicular to the x axis not the hypotenuse. Also I combined the variables as one because the height is equal to the base. Also, we weren't taught thickness so that information is not needed. However , in my problem i realize i forgot to plug in X^2

So redoing my work it would be :
V=x^5/10 from 0 to 3 which is equal to 243/10 , therefore your answer should agree with mine after you integrate your work. I'm just not sure where the thickness part come in since i have not learned that yet