# 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?

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 isbegin{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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