Double integral with three variables

Question

I have a difficulty calculating the volume of a solid that is between $$z=0,y=0,x=0, x+y+z=2 , y^2=1-z$$ . How can I work this with double integral cause draw this its difficult.

Answer 1

Warning: Solution assumes diagram is drawn but OP is encouraged to draw it out, it's not too hard as long as you go step-by-step.

The best way to do this is to draw it so that the $yz$-plane is parallel to your piece of paper. Then draw your $x+y+z = 2$ plane. After that, draw more parabolas, $z = 1-y^2$, at different points on the $x$-axis with $x \geq 0$. Try to imagine a long mountain ($z = 1- y^2$) getting cut off by the plane, $ x+y+z = 2$.

Once you've got your diagram, use ideas of 'slices' (learnt from double integrals) to construct a triple integrals. This is how I view triple integrals, i.e. extend ideas from double integrals.

Imagine you are chopping a cucumber, slice by slice with a sharp knife. Then triple integrals is can be viewd like this:

$$\underbrace{\int \int}_{\text{area of each slice}} \overbrace{\int}^{\text{how far you cut}}. \tag{1}$$

Similarly, $$\overbrace{\int}^{\text{how far you cut}}\underbrace{\int \int}_{\text{area of each slice}}. \tag{2}$$

With a diagram, this construction is extremely easy. The area of each slice can be obtained from computing the double integral of the parabola, $$\int_0^1 \int_0^{1-z} dydz.$$ How far we cut is where the long mountain generated by $z = 1- y^2$ is cut, $$\int_0^{2 - y - z} dx.$$

Use $(1)$, reason why is left for you ponder, to yield $$\int_0^1 \int_0^{1-z} \int_0^{2-y-z} dydz dx = \dots = \frac{49}{60}.$$

Computation verified via Wolfram Alpha Triple Integral widget.

Remark: This is ONE way to find the volume through integration, there are many variations to this method.