Recent content by GelatinousFur

Thanks, you are correct. The correct integral is: int(t^3*sqrt(1+4*t^2),t,0,2) So here's where I use integration by parts, but when I integrate sqrt(1+4*t^2) I have to go to an integral table. I punched int(t^3*sqrt(1+4*t^2),t,0,2) into MATLAB and it spits this answer out...

Ah, I see. Thanks for the help (and the welcome)! So then... X(t) is constant, so x(t)=1 y(t) =t z(t) = t^2 when t goes from 0 to 2. The line integral would then become: int(y^3,s) over the curve C, because z=y^2. =int(t^3*sqrt(0^2+1^2+(2t)^2),t,0,2) =int(t^3*2t,t,0,2) =2*int(t^4,t,0,2)...

Homework Statement Evaluate the line integral \[ \int_c yz\,ds.\] where C is a parabola with z=y^2 , x=1 for 0<=y<=2Homework Equations A hint was given by the teacher to substitute p=t^2 , dp=(2t)dt and use integration by parts. I also know from other line integrals with respect to arc length...