Computing volume of solid

Question

Problem :

Let $A$ denotes a region bounded by $xy=3$ and $(x+y)^2=16$.

Find the volume of the solid of revolution formed by revolving region $A$ about the line $y=x$.

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Since $A$ has symmetry, I focused on partial area of $A$ in first quadrant.

And I tried to calculate center of mass and apply Pappus's centroid theorem, but the calculation is quite messy.

So, is there any nice approach which has lower calculation than my approach?

Thanks for help.

Answer 1

One approach is to apply a rotation matrix to the entire plot to make it predominantly vertical or horizontal in the transformed Cartesian plane.

A counterclockwise rotation of $\frac{\pi}{4}$ would take $(x,y)$ in the original to $(X,Y)$ in the transformed plane where:

$X = \frac 1{\sqrt 2}x - \frac 1{\sqrt 2}y$

$Y = \frac 1{\sqrt 2}x + \frac 1{\sqrt 2}y$

where I am basically applying the formula found here: Rotation Matrix

(the formula is not that difficult to derive from first principles).

Now solve for the original coordinates in terms of the transformed to get:

$x = \frac 1{\sqrt 2}(Y + X)$

$y = \frac 1{\sqrt 2}(Y - X)$

Put in the original relationships of the curve and the bounding line to get everything in terms of transformed coordinates:

$xy = 3 \implies \frac 12(Y^2 - X^2) = 3 \implies X^2 = Y^2 - 6$

and

In the first quadrant, $x + y = 4 \implies \frac 2{\sqrt 2} Y = 4 \implies Y = \sqrt 8$

Note that the $Y$-intercept of the curve is $\sqrt 6$, and the first quadrant (in the transformed $XY$-plane) intersection of the bounding horizontal line with the curve has transformed coordinates $(\sqrt 2, \sqrt 8)$. These are all quite trivial to calculate.

You've now transformed the problem into the fairly simple integral:

$\pi\int_{\sqrt 6}^{\sqrt 8} X^2dY = \pi\int_{\sqrt 6}^{\sqrt 8} (Y^2 - 6) dY = \pi(4\sqrt 6 - \frac{10}{3}\sqrt 8)$

which can be doubled to cover all quadrants to get: $\pi(8\sqrt 6 - \frac{20}{3}\sqrt 8)$ as the required volume.

Answer 2

Change of variable can make the work easier.

$x - y = X, x + y = Y$

Jacobian for change of variable $|J| = \dfrac 12$

$xy = 3 \implies Y^2 - X^2 = 12$

$(x+y)^2 = 16 \implies Y = \pm 4$

$y = x \implies X = 0$

So the transformed region in the new coordinate system is bound between hyperbola $Y^2 - X^2 = 12$ and lines $Y = \pm4$, rotated about $X = 0$ ($Y$-axis). Using disc method, the volume element of rotated solid is,

$dV = \pi ~X^2 ~|J|~dY= \pi ~(Y^2 - 12) ~|J| ~ dY$

Note $\sqrt{12} \leq Y \leq 4~$ for the part of the solid above $Y = 0$. We have the same volume bound below $Y = 0$.