Recent content by Sebs0r

Right -> I mistook the x^2 for an x :p. Trying to do too many steps in your head inevitably leads to errors. So the new integral then is \int^{1}_{0} x^3 - 4x^2 + 4x - x^5 dx 3/4 Should be right :p Thanks man

Thanks, I just found that if I split the integral up into the first circle and two segments of the other circle, it ends up giving me a non-zero answer (not sure if right thought) Area within first quadrant = pi/4 Area in 2nd and 4th quadrant = \int^{0}_{-pi/4}\int^{cos t + sin t}_{0} r dr dt...

Homework Statement Find area bounded by x^2 + y^2 = 1 and x^2 + y^2 = x + y Homework Equations The Attempt at a Solution from the second circle, we can see r^2 >= r cos t + r sin t so r >= cos t + sin t Limits are: cos t + sin t <= r <= 1 -pi/4 <= t <= 3pi/4 Doing the...

Homework Statement Find the volume of the solid bounded by z = 0 and z = 2xy, lying in the first quadrant and bounded by the curves y = x^2 and x+y = 2 Homework Equations The Attempt at a Solution I have an answer, but just asking if I've done it correctly, since we arent given the...