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Use cylindrical coordinates to find the volume of the solid using triple integrals

Mathematics Asked by Eric Brown on December 12, 2020

The given rectangular equations are
$$x^2+y^2+z^2=64$$
$$(x-4)^2+y^2=16$$
Converting to cylindrical coordinates I get
$$r^2+z^2=64$$
$$r=4cos(theta)$$
So my triple integral is
$$4int_0^{frac{pi}{2}}int_0^{4cos(theta)}int_0^{sqrt{16-r^2}}rdzdrdtheta$$
It’s a long and tedious calculation but I worked through it and got $-frac{1408pi}{15}$ which doesn’t make sense because you can’t have a negative volume. The negative came from the second integral which is
$$4int_0^{frac{pi}{2}}int_0^{4cos(theta)}rsqrt{16-r^2}drdtheta$$
where I used a u-sub, where $u=16-r^2$ and $du=-2r$ and moving that $-frac{1}{2}$to the outside gives me
$$-2int_0^{frac{pi}{2}}int_0^{4cos(theta)}sqrt{u}dudtheta$$
So could I just swap the limits of integration to get rid of it or have I messed up somewhere else?

2 Answers

There are two mistakes I notice. You have written wrong bound of $z$ and the equation of cylinder that was already pointed out. The correct integral should be

$V = 4displaystyle int_0^{frac{pi}{2}}int_0^{8cos(theta)}int_{0}^{sqrt{64-r^2}}rdzdrdtheta approx 617.22$

$V = 4displaystyle int_0^{frac{pi}{2}}int_0^{8cos(theta)} r , sqrt{64-r^2} , dr , dtheta$

On substition, $64-r^2 = u, rdr = -frac{1}{2}du$.

The bound $r = 0$ becomes $u = 64$, $r = 8 cos theta$ becomes $u = 64 sin ^2 theta$.

$V = -2displaystyle int_0^{frac{pi}{2}}int_{64}^{64sin^2(theta)} sqrt u , du , dtheta = 2int_0^{frac{pi}{2}}int_{64sin^2(theta)}^{64} sqrt u , du , dtheta$

$V = displaystyle frac{2048}{3} int_0^{frac{pi}{2}} (1-sin^3 theta) , dtheta = frac{2048}{3} (frac{pi}{2} - frac{2}{3})$

Correct answer by Math Lover on December 12, 2020

Continuing @MathLOver's calculation, the volume is$$begin{align}4int_0^{pi/2}dthetaint_0^{8costheta}rsqrt{64-r^2}dr&=4int_0^{pi/2}dthetaleft[-frac13(64-r^2)^{3/2}right]_0^{8costheta}\&=frac{2^{11}}{3}int_0^{pi/2}dtheta(1-sin^3theta)\&=frac{2^{11}}{3}int_0^{pi/2}dtheta(1-sintheta+sinthetacos^2theta)\&=frac{2^{11}}{3}left[theta+costheta-frac13cos^3thetaright]_0^{pi/2}\&=frac{1024}{9}left(3pi-4right)\&approx617.22.end{align}$$

Answered by J.G. on December 12, 2020

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