Is it wrong to use the disk method here and use "integral" ( x^3+1)dx as my radius and multiplying it by pi and evaluating it from -1 to 1. I'm getting the correct answer when I use this method which
So basically I set the integral up like this using your method integrating from -1 to 1.
(X^3)^2-(-1)^2 and when I takin the anti derivative I get pi((x^7)/7)-x after I evaluate I get -6pi /7. I'm getting the right answer using my method.
Homework Statement
Find the volume of the solid rotated about the given axis
Homework Equations
∫R^2-r^2dx
Disk method
The Attempt at a Solution
I'm having trouble finding the limits of integration: here's my setup
Pi∫(x^3+1)dx and integrated from -1 to 1. I got this by setting...
i used the disk method here
and this is how i set up my integral
pi(x^3)^2dx
x^6 dx pi[x^(7)/7]=(pi/7)-(-pi/7) therefore giving my volume which is 2pi/7.
the books says using the sketch show how to approximate the volume of the solid by a riemann sum, hence find the volume. also, i drew a sketch and having trouble coming up with the radius and height. maybe the shel method is not as practical as others in this problem. if you could please show...
Volumes of revolution
Homework Statement
y=x^3 ; x=1; y=-1 axis:y=-1
Homework Equations
shell method ∫ 2pi*y*g(y)
The Attempt at a Solution
2pi *∫(y^(1/3)+1)*(1+y^(1/3))dy
limits of integration are from -1 to 1 i think
please help
im in calculus 2 right now and we are doing differential equaitons right now. I am confused as to why when i find the integrating factor I(x)=e^(∫p(x) and when i multiply both sides i get
e^∫(p(x))[(dy/dx)+p(x)*y]=d(e^(p(x)dx)*y) how are they equal. i will give an example.
(dy/dx)+y=x*e^(x)...