- #1

rock.freak667

Homework Helper

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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(y

^{2}) can't be expressed in terms of elementary functions. Is the reason or is there some other factor that prohibits this?