Integral over a parallelogram

Use the transformations
$$
x=3s+t : \text{ and } : y=s+2t
$$
to evaluate
$$\iint_P \left(x+2y\right)\,dA$$
over the parallelogram $P$ with vertices $(0,0)$, $(3,1)$, $(5,5)$, and $(2,4)$.

Comments

  • edited April 30

    The Jacobian

    We'll need to compute the Jacobian, that is, the determinant of

    %%[[dx//ds,dx//dt],[dy//ds,dy//dt]].%%

    Here's how:

    %%dx/(ds)=3, dx/dt=1, dy/ds=1, dy/dt=2%%

    %%[[3,1],[1,2]]=6-1=5%%

    Finding the bounds of integration

    Next, we'll find the bounds of integration:

    %%y=2x-5, y=2x, y=1/3x+10/3, y=1/3x%%

    %%s+2t=2(3s+t)-5 => s=1%%
    %%s+2t=2(3s+t) => s=0%%
    %%0<=s<=1%%

    %%s+2t=1/3(3s+t)+10/3 => t=2%%
    %%s+2t=1/3(3s+t) => t=0%%
    %%0<=t<=2%%

    Evaluating the integral

    Finally, we evaluate the integral:

    %%int_0^2int_0^1((3s+t)+2(s+2t))5dsdt=25int_0^2int_0^1(s+t)dsdt%%

    %%25int_0^2((s^2)/2+st)|_0^1dt=25int_0^2(1/2+t)dt%%

    %%25(1/2t+(t^2)/2)|_0^2=75%%

    mark
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