Find the volume V of the solid under the surface z=4-x^2-y^2 and over the rectangle R consisting of all points (x,y) such that 0<=x<=1 and 0<=y<=2.
I have started, but am unsure if my approach is correct or not.
x = 4-x^2-y^2
\int^{2}_{0}\int^{1}_{0} 4-x^{2}-y^{2} dx dy
is this correct?
i'm not sure if i can use the formula to help the integration because i also have this formula somewhere, but i don't see how else i can integrate without it as the solution with the hyperbolic substitution is too difficult for me.
nevermind i see how you did this, and i would have never known...
i found examples in class where it is integrating the double integral w.r.t dr dtheta
so do i rearrange for a to get rid of a... or do as u say and integrate w.r.t dtheta
i still have not found the limits of integration yet, i dont see how i am suppose to figure all this stuff out with no...
okay, i have read this about 10 times trying to understand this, i have tried substituting everything in to get the antiderivative you got, but i constantly get something different.
why is this integral so hard , am i doing it the hard way is there an easier way to compute this.
okay so i can only put it in polar coordinates, but where are the limits of integration in which i am suppose to integrate??
(r^2)^2 = 2a^2 (rcos^2theta - rsin^2theta) r
r^4 = 2a^2 (rcos^2theta - rsin^2theta) r
r^3 = 2a^2 (r^2cos^2theta - r^2sin^2theta)
r = 2a^2 (cos^2theta - sin^2theta)
r =...
okay, so I have done everything and integrated the whole thing, now because I have never learnt about the hyperbolic function, i am having trouble doing this.
here is my work for the rest of the integration:
(previous work)
integ (sqrt (1+x^2)dx)
let x = sinh t
dx/dt = cosh t
integ...
Evaluate \int\int_{Q}\left(1 - x^{3}\right)y^{2} dA where Q is the region bounded by y=x^2 and x = y^2
So I have drew the graphs of y=x^2 and x=y^2 and found that they intersect at (0,0) and (1,1). Now I am confused what to replace Q with, but I think it should be this: please tell me if I am...
Find the area inside the lemniscate ,which is described by the equation (x^2 + y^2) = 2a^2 (x^2 - y^2)
I have no idea where to start so I found out what a lemniscate is.. but this didn't help me much. I have rearranged the equation for a:
a = sqrt ( (x^2 + y^2)^2 / 2(x^2 - y^2) )
a = x^2...
I have tried to do the hyperbolic trig way because I am stuck on mine and I can't seem to integrate the sec^3q
this is what i have for hyperbolic, i am stuck and can't seem to integrate the last one
integ (sqrt (1+x^2)dx)
let x = sinh t
dx/dt = cosh t
integ (sqrt (1+sinh^2t) (cosh t dt))...
I can't seem to figure out how to integrate this.
integ ( sqrt (1+x^2) dx )
I am pretty sure I need to substitute the inside but I cannot figure out what. I have tried substituting x = tan theta, and i get stuck at integral (sec (theta)^3 dtheta)
here is my work: (I am goin to use q as...