For a Unit Disk

$x^2 + y^2 \ge 1$

And for a function of $x$ and $y$

$f(x, y) = 3 + y - x^2$

I want to find the volume underneath the function bound by the unit disk.

At first I integrated the function with respect to $y$, and made the upper bound $\sqrt{1 - x^2}$ and made the lower bound $-\sqrt{1 - x^2}$. I did this using Pythagoras' theorem to express the upper and lower bounds of the circle in terms of $x$. The resulting expression is $6\sqrt{1-x^2}$

Then I have to integrate the expression I just found with respect to $x$, and the upper and lower bounds are $1$ and $-1$ respectively.

However, the expression given is as follows:

The total volume

I do not understand why the expression has changed.

The expression that I worked out $6\sqrt{1-x^2}$ represents the area of a slice with a constant $x$ value, and surely I just integrate this expression with respect to x between the bounds of $1$ and $-1$ to find the total volume.


The volume is, by definition:


(**) Explanation: we have

$$\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(3+y-x^2)dydx=\int_{-1}^1\left(\left.(3-x^2)y+\frac12y^2\right)\right|_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dx= etc.$$

  • $\begingroup$ thank you. so the indefinite integral of (3+y-x^2) with respect to y is equal to: (3y)+(y^2/2) + (-yx^2) $\endgroup$ – lgdl.y Dec 16 '16 at 17:10
  • $\begingroup$ where I went wrong was thinking that this integral was only equal to 6sqrt(1-x^2) $\endgroup$ – lgdl.y Dec 16 '16 at 17:11
  • $\begingroup$ @Igd Yes to your first comment. $\endgroup$ – DonAntonio Dec 16 '16 at 17:18

Note that

$$\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(3+y-x^2)dy=\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}ydy+\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(3-x^2)dy.$$ Now

$$\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}ydy=0$$ and $$\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(3-x^2)dy=2(3-x^2)\sqrt{1-x^2}.$$

Thus the volume is given by

$$\int_{-1}^1 2(3-x^2)\sqrt{1-x^2}dx,$$ which is the expression you have.


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