Recent content by Swasse

So the range for x should be from -1 to 1? That makes sense :)

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).

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...

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...

Just figured it out, the radius is not 9, it is √9=3. Is that the correct limit?

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?

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.

My mistake, it should have said 1<= x^2+ 9y^2<= 9

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...