∫∫_{D} x^{2}y dA where D is the region between y=sqrt x, y= 1/x and x=3
∫∫∫_{E} x dV where E is the volume bounded by z = 4 - x^2 - y^2 and z = x^2+3y^2 and is in the first octant (x>=0, y>=0, z>=0)
Thanks for the help
Start by drawing a sketch.
It is easiest to evaluate this integral using horizontal strips. You can see that the region is bounded on the left by the y = 1/x graph, but since we are integrating over x first, that means it's bounded by x = 1/y, and the region is bounded on the right by the y = sqrt(x) graph, but since we are integrating over x first, that means it's bounded by x = y^2.
We can also see that the horizontal strips are bounded below by y = 1 and bounded above by y = 3.
So $\displaystyle \displaystyle \begin{align*} \int{\int_D{x^2y}\,dA} = \int_1^3{\int_{\frac{1}{y}}^{y^2}{x^2y\,dx}\,dy} \end{align*}$