Triple Integral (Type 1 Solid Region) (example continued)

The first integration (for Type 1 Solid Region) is in the z-direction.

z = 1 – y

I = ∫ ∫ ∫ [ f dz ] dA

z = 0

Now the element of area, dA, may be either dx dy or dy dx (depending on the order

of integration). So determine the limits of integration in the x and y – directions by

projecting the solid region on to the xy-plane as shown below.

Note: The limits of integration on the inner integral contain at most two variables, the

limits of integration on the middle integral contain at most one variable, and the limits

of integration on the outer integral must be constants.

Option 1: By inspection of the intersections, the limits of integration in the x-direction

(figure on the left) are x = 0 to x = y² . Next “sweep” the element of area in the y-direction.

In this case the limits of integration are from y = 0 to y = 1. So the integral becomes

y=1 x= y² z = 1 – y

I = ∫ ∫ ∫ [ f(x,y,z) dz ] dx dy (result)

y=0 x=0 z = 0

Option 2: By inspection of the intersections, the limits of integration in the y-direction

(figure on the right) give y = √x to y = 1 . Then “sweep” the element of area in the x-direction.

In this case the limits of integration are from x = 0 to x = 1. So the integral becomes

x=1 y = 1 z = 1 – y

I = ∫ ∫ ∫ [ f(x,y,z) dz ] dy dx (result)

x=0 y=√x z = 0

Click here for a Type 2 Solid Region example. Click here for a Type 3 Solid Region example.

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