Finding volume for triple integrals in cylindrical coordinates — Krista King Math

15.7: Triple Integrals in Cylindrical Coordinates
15.7: Triple Integrals in Cylindrical Coordinates

Finding volume for triple integrals in cylindrical coordinates

Converting to cylindrical coordinates

We can use triple integrals and cylindrical coordinates to solve for the volume of a solid cylinder. The volume formula in rectangular coordinates is

???V=\int\int\int_Bf(x,y,z)\ dV???

where ???B??? represents the solid cylinder and ???dV??? can be defined in cylindrical coordinates as

???dV=r\ dz\ dr\ d\theta???

To convert in general from rectangular to cylindrical coordinates, we can use the formulas

???x=r\cos{\theta}???

???y=r\sin{\theta}???

???z=z???

and

???r^2=x^2+y^2???

Remember, rectangular coordinates are given as ???(x,y,z)???, and cylindrical coordinates are given as ???(r,\theta,z)???.

Finding volume from a triple integral in cylindrical coordinates

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Volume in cylindrical coordinates of the solid in the cylinder, above the plane, and below the cone

Example

Use cylindrical coordinates to find the volume of the triple integral, where ???E??? is the solid that lies within the cylinder ???x^2+y^2=4???, above the plane ???z=0??? and below the cone ???z^2=9x^2+9y^2???.

???\int\int\int_E3x^2\ dV???

First, we’ll convert the function we were given into cylindrical coordinates using the conversion formulas.

???3x^2???

???3\left(r\cos{\theta}\right)^2???

???3r^2\cos^2{\theta}???

Replacing the original function with this one, we get

???\int\int\int_E3r^2\cos^2{\theta}\ dV???

Substituting for ???dV???, the integral becomes

???\int\int\int_E3r^2\cos^2{\theta}\left(r\ dz\ dr\ d\theta\right)???

???\int\int\int_E3r^3\cos^2{\theta}\ dz\ dr\ d\theta???

Now we just need to find limits of integration. We’ve been told that we’re interested in the solid that lies inside the cylinder ???x^2+y^2=4???. If we convert this to cylindrical coordinates using ???r^2=x^2+y^2???, we get

???r^2=4???

???r=2???

Since ???r??? represents radius, and radius can only be positive, we can say that the limits of integration for ???r??? are ???[0,2]???, and therefore

???\int\int_0^2\int3r^3\cos^2{\theta}\ dz\ dr\ d\theta???

We’ve also been told that the solid lies above the plane ???z=0???. Since no conversion is required for ???z??? when we’re moving from rectangular to cylindrical coordinates, we can leave this as-is. We also know that the solid lies below ???z^2=9x^2+9y^2???. We’ll factor out ???9??? and get ???z^2=9\left(x^2+y^2\right)???. Using ???r^2=x^2+y^2??? to convert this to cylindrical coordinates, we get

???z^2=9\left(x^2+y^2\right)???

???z^2=9\left(r^2\right)???

???z^2=9r^2???

???\sqrt{z^2}=\sqrt{9r^2}???

???z=3r???

Putting these two pieces of information together, we can say that the limits of integration for ???z??? are ???[0,3r]???, and therefore

???\int\int_0^2\int_0^{3r}3r^3\cos^2{\theta}\ dz\ dr\ d\theta???

For all full cylinders, the limits of integration for ???\theta??? will be ???[0,2\pi]???, therefore

???\int_0^{2\pi}\int_0^2\int_0^{3r}3r^3\cos^2{\theta}\ dz\ dr\ d\theta???

We can use triple integrals and cylindrical coordinates to solve for the volume of a solid cylinder.

We always integrate inside out, so we’ll integrate with respect to ???z??? first, treating all other variables as constants.

???\int_0^{2\pi}\int_0^23r^3z\cos^2{\theta}\Big|_{z=0}^{z=3r}\ dr\ d\theta???

???\int_0^{2\pi}\int_0^23r^3(3r)\cos^2{\theta}-3r^3(0)\cos^2{\theta}\ dr\ d\theta???

???\int_0^{2\pi}\int_0^29r^4\cos^2{\theta}\ dr\ d\theta???

Now we’ll integrate with respect to ???r???, treating all other variables as constants.

???\int_0^{2\pi}9\left(\frac15\right)r^5\cos^2{\theta}\Big|_{r=0}^{r=2}\ d\theta???

???\int_0^{2\pi}\frac95r^5\cos^2{\theta}\Big|_{r=0}^{r=2}\ d\theta???

???\int_0^{2\pi}\frac95(2)^5\cos^2{\theta}-\frac95(0)^5\cos^2{\theta}\ d\theta???

???\int_0^{2\pi}\frac{288}{5}\cos^2{\theta}\ d\theta???

We’ll make a substitution using the trigonometric identity

???\cos^2{\theta}=\frac12\left[1+\cos{(2\theta)}\right]???

So the integral becomes

???\int_0^{2\pi}\frac{288}{5}\left\{\frac12\left[1+\cos{(2\theta)}\right]\right\}\ d\theta???

???\int_0^{2\pi}\frac{144}{5}\left[1+\cos{(2\theta)}\right]\ d\theta???

???\int_0^{2\pi}\frac{144}{5}+\frac{144}{5}\cos{(2\theta)}\ d\theta???

???\frac{144}{5}\theta+\frac{144}{5(2)}\sin{(2\theta)}\Big|_0^{2\pi}???

???\frac{144}{5}\theta+\frac{72}{5}\sin{(2\theta)}\Big|_0^{2\pi}???

???\frac{144}{5}(2\pi)+\frac{72}{5}\sin{\left[2(2\pi)\right]}-\left[\frac{144}{5}(0)+\frac{72}{5}\sin{\left[2(0)\right]}\right]???

???\frac{144}{5}(2\pi)+\frac{72}{5}\sin{(4\pi)}-\frac{72}{5}\sin{0}???

???\frac{288\pi}{5}+\frac{72}{5}(0)-\frac{72}{5}(0)???

???\frac{288\pi}{5}???

This is the volume of the solid.

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