# Recent content by Swasse

1. ### Triple Integral

So the range for x should be from -1 to 1? That makes sense :)
2. ### Triple Integral

Range or volume of integration might be more correct, English is not my first language so I hope you understand anyway. The problem just says to evaluate the integral over the set D={(x,y,z)∈R^3 : 0<=z<=y<=x^2<=1}. The answer should be (1/3)*(13/15 - pi/4).
3. ### Triple Integral

What I meant with 0<=z<=y<=x^2<=1, was that the area of integration is D, where D={(x,y,z)∈R^3 : 0<=z<=y<=x^2<=1}. Does that make more sense? Anyway, I tried to evaluate your integral and got (1/6)*(13/15 - pi/4), but it should be twice as much according to my key. Your limits seems correct...
4. ### Triple Integral

Homework Statement ∫∫∫z/(1+x^2)dxdydz with the range 0<=z<=y<=x^2<=1 Homework Equations The Attempt at a Solution I have tried with the limits 0<=z<=1; z<=y<=1 and sqrt(y)<=x<=1, but it doesn't get me the right answer. Can you please help me and maybe give me a step-by-step...
5. ### Double integral substitution

Just figured it out, the radius is not 9, it is √9=3. Is that the correct limit?
6. ### Double integral substitution

I had a feeling it was, because my answer was in the right form, just too large. Can you please explain why it is wrong?
7. ### Double integral substitution

Can someone please confirm if my way of solving is correct? I made a typo in the first post, but I have corrected it now.
8. ### Double integral substitution

My mistake, it should have said 1<= x^2+ 9y^2<= 9
9. ### Double integral substitution

Hey. Homework Statement ∫∫x^3 dxdy, with the area of integration: D={(x,y)∈R^2: 1<=x^2+9y^2<=9, x>=3y} The Attempt at a Solution Did the variable substitution u=x and v=3y so the area of integration became 1<=u^2 + v^2 <=9, u>=v. And the integral became ∫∫(1/3)u^3 dudv. Then I switched to...