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This is a question from a textbook I am reading:

Find the volume when the region R is revolved about the y-axis when R is defined as the region bounded by $x = \sqrt{16-y^2}$ and $x = 0$.

Because the function is symmetrical about the y-axis and being revolved about the y-axis, why don't these two integrals evaluate to the same value?

Integral one: $$\int_0^4\!\pi\left(\sqrt{16-y^2}\right)^2\,dy = \frac{128\pi}3$$

Integral two: $$\int_{-4}^4\!\pi\left(\sqrt{16-y^2}\right)^2\,dy = \frac{256\pi}3$$

It seems to me that they represent the same volume--like creating a semicircular volume by revolving a fourth of a circle about its base, or by revolving a semicircle along its base.

I think that the second equation overestimates the volume because it counts the same volume twice. (The textbook says otherwise!)

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  • $\begingroup$ The second integral is the double of the first one since it computes the first one two times. $\endgroup$ – Sigur May 17 '13 at 22:50
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    $\begingroup$ $\displaystyle\int^a_{-a} f(t)dt = 2\int^a_0 f(t)dt$ if $f$ is even. $\endgroup$ – Shuhao Cao May 17 '13 at 22:51
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The function is not symmetrical about the $y$-axis, indeed the curve "lives" in the first and fourth quadrants.

There is symmetry about the $x$-axis, which is why one integral is twice the other.

Remark: Since you are much more familiar with integration with respect to $x$, you might interchange the roles of $x$ and $y$ (geometrically: reflect in the line $y=x$). Then you will see more clearly what is going on. The first integral gives the volume of a half-sphere. The second gives the volume of a full sphere.

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  • $\begingroup$ Thank you, when I visualized the graph I mistakenly used the usual x-y Cartesian plane as a background, forgetting to consider the format of the bounding equation. $\endgroup$ – d0rmLife May 17 '13 at 23:05
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    $\begingroup$ Yes, they pulled a switch on you. Not nice! $\endgroup$ – André Nicolas May 17 '13 at 23:11

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