double integration help

[j2009t]

1. Sketch the region of integration & reverse the order of integration. Double integral y dydz... 1st (top=1, bottom =0)... 2nd(inner) integral (top=cos(piex), bottom=(x-2)...

2. Evaluate the integral by reversing the order of integration. double integral sqrt(2+x^3) dxdy... 1st integral (top=1, bottom=0)... 2nd (interior) integral(top=1, bottom=sqrt(y)).

3. Evaluated the integral using a change of variables; double integral sin(x+y) dxdy... below both integral signs is R... over the region R being on the disc x^2 +u^2<=2.

Thanks!

 

[galactus]

1. Sketch the region of integration & reverse the order of integration. Double integral y dydz... 1st (top=1, bottom =0)... 2nd(inner) integral (top=cos(piex), bottom=(x-2)...

I assume you mean \(\displaystyle \int_{0}^{1}\int_{x-2}^{cos(\pi x)}ydydx\). That z is a typo?.

Anyway, to reverse the order, we have:

\(\displaystyle y=x-2\)

\(\displaystyle y=cos(\pi x)\)

\(\displaystyle x=0\)

\(\displaystyle x=1\)

Therefore, the new limits become:

\(\displaystyle x=y+2\)

\(\displaystyle y=\frac{cos^{-1}(y)}{\pi}\)

\(\displaystyle x=-1\)

\(\displaystyle x=1\)

See why?. The new y limits are obtained by subbing the old x limits into \(\displaystyle cos(\pi (1))=-1, \;\ cos(\pi(0))=1\)

\(\displaystyle \int_{-1}^{1}\int_{y+2}^{\frac{cos^{-1}(y)}{\pi}}ydxdy\)