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Homework Statement
1) By evaluating a double integral find the volume of the tetrahedron bounded by planes x+2y+z=2, x= 2y ,z=0 and y=0 (answer = 1/3)
2)
[tex]\int_{0} ^{1} \int_{x} ^{1} sin(y^2) dydx[/tex]
i) explain why this can be evaluated by integrating wrt y first.
ii) sketch the region of integration R of the double integration
iii)Evaluate the integral by expressing it as a Type II region first.
Homework Equations
The Attempt at a Solution
My query is with 1 and 2ii)
for 1
my region I integrated over was x/2≤y≤(2-x)/2 and 0≤x≤1
found when z=0 to get y=(2-x)/2 and this intersects y=x/2 at (1,1/2)
yet when I compute
[tex]\int_{0} ^{1} \int_{\frac{x}{2}} ^{\frac{2-x}{2}} (2-2y-x) dydx[/tex]
I don't get 1/3. If it's a problem with my algebra I'll fix it. But are my boundaries correct?
2) I believe it can't be done with dy first since the integral of sin(y2) can't be expressed in terms of elementary functions. Is the reason or is there some other factor that prohibits this?