Question

Evaluate the triple integral
∫∫∫xyzdV where E is the solid 0<=z<=1, 0<=y<=z, 0<=x<=y.
I understand to plug in the E's for the limits of integration but I keep winding up with a variable in my answer, and it needs to be a number.

Answer 1

what order are you doing the integrals?

Answer 2

dxdydz

Answer 3

\[\int\limits_{0}^{1}dz \int\limits_{0}^{z}dy \int\limits_{0}^{y}dx (xyz)\]Since there is another integral other than the one in respect to z that uses this variable we cant integrate that first, the same happens with y, so we need to integrate in respect to x first.\[\int\limits\limits_{0}^{1}dz \int\limits\limits_{0}^{z}dy \int\limits\limits_{0}^{y}dx (xyz)=\int\limits\limits_{0}^{1}dz \int\limits\limits_{0}^{z}dy \left[\frac{x^2yz}{2}\right]_{0}^{y}=\int\limits\limits_{0}^{1}dz \int\limits\limits_{0}^{z}dy \frac{y^3z}{2}\]Now z is still being useg in another integral so we integrate in respect to y:\[\int\limits\limits\limits_{0}^{1}dz \int\limits\limits\limits_{0}^{z}dy \frac{y^3z}{2}=\int\limits\limits\limits_{0}^{1}dz \left[\frac{y^4z}{8}\right]_{0}^{z}=\int\limits\limits\limits_{0}^{1}dz \frac{z^5}{8}=\left[\frac{z^6}{48}\right]_{0}^{1}=\frac{1}{48}\]

Answer 4

dx converts all x's to y's dy will convert all y's to z's and then like ivan posted 1/48