Calculus Definite Integrals: Volumes by Washer Method

[jsun2015]

Homework Statement

Using Washer Method: Revolve region R bounded by y=x^2 and y=x^.5 about y=-3

Homework Equations

V= integral of A(x) from a to b with respect to a variable "x"
A(x)=pi*radius^2

The Attempt at a Solution

pi(integral of (x^.5-3)^2 -(x^2)^2-3) from 0 to 1 with respect to x

The answer involves x^.5+3 instead of x^.5 -3. I don't understand why you would add instead of subtract 3; y decreased from 0 to -3 when revolved around y=-3

 

[BvU]

Make a drawing. See that the outer radius of your washers is $\sqrt x + 3$ and the inner is $x^2 + 3$

 

[LCKurtz]

The answer involves x^.5+3 instead of x^.5 -3. I don't understand why you would add instead of subtract 3; y decreased from 0 to -3 when revolved around y=-3

The length of a vertical line is $y_{upper}-y_{lower}$. Also don't forget to square your second term.