Finding Volumes and Evaluating Double Integrals

[rock.freak667]

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)

\int_{0} ^{1} \int_{x} ^{1} sin(y^2) dydx

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

\int_{0} ^{1} \int_{\frac{x}{2}} ^{\frac{2-x}{2}} (2-2y-x) dydx

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?

 

[LCKurtz]

Your boundaries look correct in your integral for 1.

And your answer to 2 looks correct too. And that problem will go away when you reverse the integration order.

 

[rock.freak667]

ok for some reason I can't seem to get 1/3. This is what I have


\int_{0} ^{1} \int_{\frac{x}{2}} ^{\frac{2-x}{2}} (2-2y-x)dydx

= \int_{0} ^{1} \left[ 2y-y^2 -xy \right] _{\frac{x}{2}} ^{\frac{2-x}{2}} dx

= \int_{0} ^{1} \left( (2-x)- \frac{1}{4}(2-x)^2 -\frac{x}{2}(2-x) -x + \frac{x^2}{4} + \frac{x^2}{2} \right)


= \int_{0} ^{1} (2-x-1+x-\frac{x^2}{4}-x+\frac{x^2}{2}-x+\frac{x^2}{4}+\frac{x^2}{2}) dx

= \int_{0} ^{1} (1-2x+\frac{x^2}{2}) dx

= \left[ x-x^2 +\frac{x^3}{6} \right]_{0} ^{1} = 1/6

where am I going wrong? Because I clearly can't see it

EDIT: for the second one as well, I can't get the answer, I must be doing something wrong

\int_{0} ^{1} \int_{x} ^{1} siny^2 dxdy

=\int_{0} ^{1} \left[ xsiny^2 \right] _{x} ^{1} dy

=\int_{0} ^{1} (xsin1-xsinx^2) dx

= \left[ \frac{x^2}{2}sin1 + \frac{1}{2}cos(x^2) \right] _{0} ^{1}

= \frac{1}{2}sin1 + \frac{1}{2}cos1 - \frac{1}{2}


answer is (1-cos1)/2

 

[LCKurtz]

It's late here so I will be brief. Your limits are correct on the first one and Maple checks it as 1/3. Recheck your algebra.

On the second one your reversed limits are wrong. You need to draw a picture of the region from the first integral and use it to get the dx dy limits.

Sack time here.

 

[rock.freak667]

Thanks for your help, I just need a bit of understanding with the 2nd one still. I found the new limits as y≤x≤0 and 0≤y≤1, but to get the answer the y range must be 1≤y≤0 and this makes no sense to me.

 

[LCKurtz]

Thanks for your help, I just need a bit of understanding with the 2nd one still. I found the new limits as y≤x≤0 and 0≤y≤1, but to get the answer the y range must be 1≤y≤0 and this makes no sense to me.

Agreed, 1≤y≤0 makes no sense. Again, to work the problem you must draw the region described in the original integral. You should see a triangular region in the first quadrant above the line y = x and below the line y = 1. Then simply use that picture to get the limits integrating in the x direction first. You need to look at the picture to get the new limits.

 

[rock.freak667]

Agreed, 1≤y≤0 makes no sense. Again, to work the problem you must draw the region described in the original integral. You should see a triangular region in the first quadrant above the line y = x and below the line y = 1. Then simply use that picture to get the limits integrating in the x direction first. You need to look at the picture to get the new limits.

I got out this one, but in my integration working above can you see any error I am making because I keep getting answers which are not 1/3 . I just tried it again and got an answer of -1/2

 

[LCKurtz]

The x²/2 term in the second line from the bottom is wrong.