Question

using a triple integral, what is the volume of the solid in the first octant bounded by the functions x^2+y^2+z^2=4, y=x, and z=y? the solution must only use cartesian coordinates.

Answer 1

here's a hint example :
Calculation of a triple integral in Cartesian coordinates can be reduced to the consequent calculation of three integrals of one variable.

Consider the case when a three dimensional region U is a type I region, i.e. any straight line parallel to the z-axis intersects the boundary of the region U in no more than 2 points. Let the region U be bounded below by the surface z = z1(x,y), and above by the surface z = z2(x,y) (Figure 1). The projection of the solid U onto the xy-plane is the region D (Figure 2). We suppose that the functions z1(x,y) and z2(x,y) are continuous in the region D.

calculating a triple integral is reduced to calculating a double integral, where the integrand is an one-dimensional integral. In the given case, we need first to calculate the inner integral with respect to the variable z, and then the double integral with respect to the variables x and y.

Answer 2

hope it helps!!

Answer 3

integrate this:
\[\int\limits_{0}^{2/\sqrt3}\int\limits_{x}^{\sqrt{2-(x^2/2)}}\int\limits_{y}^{\sqrt{4-x^2-y^2}}dV\]
to obtain the volume :D