Recent content by chevy900ss

i thought it was x=sqrt(1-y)

ok so i am not sure how to get the new boundries but its r^2rdrdtheta which integrated goes to r^4/4dtheta. Is this right?

Homework Statement Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. int(-1to1)int((sqrt(1-y^2))to(sqrt(1-y))[x^2+y^2]dxdy Homework Equations x=rcostheta y=rsintheta The Attempt at a Solution...

Integrate int(0topi)int(0tosinx)ydydx

i see i was just no where near the right answer.

\frac{1}{\pi^2}\int_0^\pi\int_0^\pi sin(x+y)dydx=\frac{1}{\pi^2}\int_0^\pi sin(x)[-cosy|_0^\pi] dx=\frac{1}{\pi^2}\int_0^\pi (sinx-1)dx=\frac{1}{\pi^2}[-cosx-x|_0^\pi]=\frac{-\pi}{\pi^2}=\frac{-1}{\pi}

\frac{1}{pi^2}\int_0^pi\int_0^pi sin(x+y)dydx=\frac{1}{pi^2}\int_0^pi sin(x)[-cosy|_0^pi] dx=\frac{1}{pi^2}\int_0^pi (sinx-1)dx=\frac{1}{pi^2}[-cosx-x|_0^pi]=\frac{-pi}{pi^2}=\frac{-1}{pi}

\frac{1}{pi^2}\int_0^pi\int_0^pi sin(x+y)dydx=\frac{1}{pi^2}\int_0^pi sin(x)[-cosy|_0^pi] dx=\frac{1}{pi^2}\int_0^pi (sinx-1)dx=\frac{1}{pi^2}[-cosx-x|_0^pi]=\frac{-pi}{pi^2}=\frac{-1}{pi}

the end integral i get 1/pi^2[-cosx-x(from 0 to pi)] which i get -1/pi

For A i get pi^2

yeah i can't figure out how to make it look like yours. It just stays in the formula form

\frac{1}{A}\int_{x = 0}^\pi \int_{y = 0}^\pi sin(x + y) dy~dx

well i don't no how to show the symbols

Homework Statement Find the average value of f(x,y)=sin(x+y) over the rectangle 0 ≤ x ≤ pi, 0 ≤ y ≤ pi. Homework Equations I need to know if this is the right answer please. The Attempt at a Solution I got 1/pi as a answer.

If anyone could help my figure it out i would really appreciate it. Thanks