Iterated integral

[steller]

Let S be the surface defined by \(\displaystyle z= f(x,y) = 1-y-x^2 \)
Let V be the volume of the 3-D region in the first octant bounded by S and the coordinate planes.

Setup the iterated integrals for V in two ways:

a) integrate first respect to x and then respect to y
b) integrate first respect to y and then respect to x

My attempt: coordinate plane boundaries is

\(\displaystyle z= 0. \)

\(\displaystyle 1-y-x^2=0 \)

So, x = 0, to 1

\(\displaystyle y = 1-x^2 \)

so, y = 0 to \(\displaystyle 1-x^2 \)

Setting it up:

\(\displaystyle \int_0^{1-x^2} \int_0^1(1-y-x^2)\mathrm{d}x \mathrm{d}y\)

Is this correct so far just setting it up?

 

[girodd]

Let S be the surface defined by \(\displaystyle z= f(x,y) = 1-y-x^2 \)
Let V be the volume of the 3-D region in the first octant bounded by S and the coordinate planes.

Setup the iterated integrals for V in two ways:

a) integrate first respect to x and then respect to y
b) integrate first respect to y and then respect to x

My attempt: coordinate plane boundaries is

\(\displaystyle z= 0. \)

\(\displaystyle 1-y-x^2=0 \)

So, x = 0, to 1

\(\displaystyle y = 1-x^2 \)

so, y = 0 to \(\displaystyle 1-x^2 \)

Setting it up:

\(\displaystyle \int_0^{1-x^2} \int_0^1(1-y-x^2)\mathrm{d}x \mathrm{d}y\)

Is this correct so far just setting it up?

​If you integrate first wrt x, then the x-limits are dependent on y. x ranges from 0 to sqrt(1-y), and y from 0 to 1.

 

[HallsofIvy]

In the xy-plane, \(\displaystyle z= 0= 1- y- x^2\) so that \(\displaystyle y= 1- x^2\) a parabola that crosses the y-axis at (0, 1) and the x-axis at (1, 0). Since we are in the first quadrant we also have that \(\displaystyle x= \sqrt{1- y}\)

If you are integrating with respect to y first, then x, x goes from 0 to 1 and, for each x, y goes from 0 to \(\displaystyle 1- x^2\).

If you are integrating with respect to x first, then y, y goes from 0 to 1 and, for each y, x goes from 0 to \(\displaystyle \sqrt{1- y}\).