Ratio Test
Calculus II, Section 3 Summary
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Example 1.

Suppose we want to evaluate

    \[ \int \int \int_{S}dV \]

where S is the solid enclosed by the planes z=1 and z=4 and the cylinders x^{2}+y^{2}=1,x^{2}+y^{2}=4. We can use the cylindrical coordinates as

    \[ x=r\cos \theta ,y=r\sin \theta ,z=z,dV=rdzdrd\theta \]

We see that the bottom of the given long washer is a region between to circles x^{2}+y^{2}=1,x^{2}+y^{2}=4 and so 1\leq r\leq 2 and 0\leq\theta \leq 2\pi . We write

    \begin{eqnarray*} \int \int \int_{S}dV &=&\int_{0}^{2\pi }\int_{1}^{2}\int_{1}^{4}rdzdrd\theta \\ &=&6\pi \int_{1}^{2}rdr \\ &=&9\pi \end{eqnarray*}

Example 2.

Suppose we want to evaluate

    \[ \int \int \int_{S}ydV \]

where S is the solid enclosed by the plane z=1,z=x+3 and the cylinders x^{2}+y^{2}=2,x^{2}+y^{2}=3.We can use the cylindrical coordinates as

    \[ x=r\cos \theta ,y=r\sin \theta ,z=z,dV=rdzdrd\theta \]

We see that the bottom of the given long washer is a region between to circles x^{2}+y^{2}=2,x^{2}+y^{2}=3 and so \sqrt{2}\leq r\leq \sqrt{3} and 0\leq \theta \leq 2\pi . We write

    \begin{eqnarray*} \int \int \int_{S}ydV &=&\int_{0}^{2\pi }\int_{\sqrt{2}}^{\sqrt{3}% }\int_{1}^{4}\left( r\sin \theta \right) rdzdrd\theta \\ &=&\int_{0}^{2\pi }\sin \theta \int_{\sqrt{2}}^{\sqrt{3}}r^{2}% \int_{1}^{4}dzdrd\theta \end{eqnarray*}

Give it a try and compute the rest.

Example 3.

Suppose we want use triple integral to find the volume of the solid enclosed by the intersection of the sphere x^{2}+y^{2}+z^{2}=4 and the cylinder x^{2}+y^{2}=2x. Using cylindrical coordinates, we know

    \[ x=r\cos \theta ,y=r\sin \theta ,z=z,dzdydx=rdzdrd\theta \]

We see that

    \[ x^{2}+y^{2}+z^{2}=4\Rightarrow z=\pm \sqrt{4-x^{2}-y^{2}}=\pm \sqrt{4-r^{2}} \]

and

    \[ x^{2}+y^{2}=2x\Rightarrow r=2\cos \theta \]

Note that -\frac{\pi }{2}\leq \theta \leq \frac{\pi }{2} and 0\leq r\leq 2\cos \theta will provide the region of integration on xy-plane. Therefore, we write

    \[ \int_{-\pi /2}^{\pi /2}\int_{0}^{2\cos \theta }2\int_{0}^{\sqrt{4-r^{2}}% }rdzdrd\theta \]

Since the solid id symmetric with respect to xy-plane, we can compute half volume and multiply by 2. Continue with the computation of this triple integral.

Example 4.

Suppose we are given

    \[ \int_{0}^{1}\int_{0}^{\sqrt{1-y^{2}}}\int_{0}^{\sqrt{1-x^{2}-y^{2}}% }e^{\left( x^{2}+y^{2}+z^{2}\right) ^{3/2}}dzdxdy \]

We want to use spherical coordinates to compute this triple integral. We start by writing

    \begin{eqnarray*} \rho ^{2} &=&x^{2}+y^{2}+z^{2} \\ x &=&\rho \sin \varphi \cos \theta \\ y &=&\rho \sin \varphi \sin \theta \\ z &=&\rho \cos \varphi \\ dzdxdy &=&\rho ^{2}\sin \varphi d\rho d\theta d\varphi \end{eqnarray*}

We see that the region of integration is a sphere of radius 1 in the first octant. Therefore, 0\leq \rho \leq 1,0\leq \theta \leq \frac{\pi }{2},0\leq\varphi \leq \frac{\pi }{2} and so

    \begin{eqnarray*} \int_{0}^{1}\int_{0}^{\sqrt{1-y^{2}}}\int_{0}^{\sqrt{1-x^{2}-y^{2}}% }e^{\left( x^{2}+y^{2}+z^{2}\right) ^{3/2}}dzdxdy &=&\int_{0}^{\pi /2}\int_{0}^{\pi /2}\int_{0}^{1}e^{\left( \rho ^{2}\right) ^{3/2}}\rho ^{2}\sin \varphi d\rho d\theta d\varphi \\ &=&\frac{\pi }{2}\int_{0}^{\pi /2}\sin \varphi \int_{0}^{1}e^{\left( \rho ^{3}\right) }\rho ^{2}d\rho d\varphi \\ &=&\frac{e-1}{6} \end{eqnarray*}

Example 5.

We would like to use spherical coordinates to integrate f\left(x,y,z\right) =\sqrt{x^{2}+y^{2}+z^{2}} over the solid above the cone z=-\sqrt{3x^{2}+3y^{2}} and inside the sphere x^{2}+y^{2}+z^{2}=4. We know the cone is heavy side down, pointing upward. The elements that we need are \rho ,\theta and \varphi . First

    \[ 0\leq \rho \leq 2 \]

since radius of the sphere is at max 2. Next, to get the whole circle on xy-plane we need

    \[ 0\leq \theta \leq 2\pi \]

To find the bounds for \varphi , ask yourself how far the angle \varphi goes from the begining point on the positive side of the z-axis to get to the cone. Consider the xz-plane only. If we travel one unit down that is if we take z=-1 then x=\pm \frac{1}{\sqrt{3}} and the obtained triangle provides that the angle that this cone makes with the z-axis is \frac{\pi }{6}. However, the range of \varphi must be 0\leq \varphi \leq \pi -\frac{\pi }{6}=\frac{\pi }{5}. We now are ready to set up the triple integral and write

    \begin{eqnarray*} \int_{0}^{\pi /5}\int_{0}^{2\pi }\int_{0}^{2}\sqrt{\rho ^{2}}\rho ^{2}\sin \varphi d\rho d\theta d\varphi &=&8\pi \int_{0}^{\pi /5}\sin \varphi d\varphi \\ &=&4\left( 2+\sqrt{3}\right) \pi \end{eqnarray*}

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