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I am trying to solve this problem:

$$2\int_{-2}^2\int_0^{\sqrt{4-x^2}}(4-y)\,dx\,dy$$

This is supposed to be the area of a cylinder defined by $x^2+y^2=4$ and cut by the plane $z=4-y$.

Is this even the proper integral to use? The bounds on the inner integral do not make sense to me.

The textbook states that this integral evaluates to $16\pi$. Is that true?

Michael Hardy
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user753407
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  • Yeah, I think the simplest fix is to rewrite $dx,dy$ as $dy,dx$. – Brian Tung Feb 23 '20 at 00:35
  • Thanks! I did do that and still did not get 16pi. Sorry, my original post was unfocused. @BrianTung – user753407 Feb 23 '20 at 01:46
  • How did you get $$2\int_{-2}^2\int_0^{\sqrt{4-x^2}}(4-y)dxdy?$$ Assuming $z=0$ is the lower plane, I get $$\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{0}^{4-y},dz,dy,dx$$ which does evaluate to $16\pi$. – Axion004 Feb 23 '20 at 04:23
  • @Axion004: Your innermost integral does in fact evaluate to $4-y$. The rest is then just symmetry plus switching $dx,dy$ for $dy,dx$. Ahh, and that is the problem for the OP; they're using symmetry across the $x$-axis, but the figure is not symmetrical across the $x$-axis; it's symmetrical across the $y$-axis. – Brian Tung Feb 23 '20 at 05:12
  • That is true. But, how does the OP know that the innermost integral evaluates to $4-y$? I found it was $4-y$ because I assumed that the OP forgot to include the plane $z=0$. Without this information, I don't know how the OP found $(4-y)$ as the body of the question doesn't explain how the OP found this. – Axion004 Feb 23 '20 at 05:36

1 Answers1

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[Note: We assume that the desired region is bounded by the given plane and by the plane $z = 0$.]

The integral in the question employs symmetry, but across the wrong axis. The figure as described is bilaterally symmetric across the $y$-axis; that is, we can divide the figure into two halves, one with $x > 0$, and the other with $x < 0$. However, the integral as written has two flaws:

  • It has $dx \, dy$ when it should have $dy \, dx$
  • Its limits imply symmetry across the $x$-axis and not the $y$-axis.

The correct integral should have been

$$ \int_{x=-2}^2 \int_{y=-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 4-y \, dy \, dx = 2 \int_{x=0}^2 \int_{y=-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 4-y \, dy \, dx $$

When evaluated, the result of this is $16 \pi$.

Brian Tung
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