By the method of double integration, we can see that the volume is the iterated integral of the form [latex]\underset{R}{\displaystyle\iint}{(1-{x^2}-{y^2})}{dA}[/latex] where [latex]{R} = {\left \{{(r,{\theta})}{\mid}{0} \ {\leq} \ {r} \ {\leq} \ {{1},{0}} \ {\leq} \ {\theta} \ {\leq} \ {2{\pi}} \right \}}[/latex].

This integration was shown before in Example “Evaluating a Double Integral by Converting from Rectangular Coordinates”, so the volume is [latex]{\frac{\pi}{2}}[/latex] cubic units.

First change the disk [latex]{(x-1)^2} + {y^2} = {1}[/latex] to polar coordinates. Expanding the square term, we have [latex]{x^2} - {2x} + {1} + {y^2} = {1}[/latex]. Then simplify to get [latex]{x^2} + {y^2} = {2x}[/latex], which in polar coordinates becomes [latex]{r^2} = {2r}{\cos}{\theta}[/latex] and then either [latex]{r} = {0}[/latex] or [latex]{r} = {2}{\cos}{\theta}[/latex]. Similarly, the equation of the paraboloid changes to [latex]{z} = {4} - {r^2}[/latex]. Therefore we can describe the disk [latex]{(x-1)^2} + {y^2} = {1}[/latex] on the [latex]xy[/latex]-plane as the region

[latex]{D} = {\left \{{(r,{\theta})}{\mid}{0} \ {\leq} \ {\theta} \ {\leq} \ {{\pi},{0}} \ {\leq} \ {r} \ {\leq} \ {2{\cos}{\theta}} \right \}}[/latex].

Hence the volume of the solid bounded above by the paraboloid [latex]{z} = {4} - {x^2} - {y^2}[/latex] and below by [latex]{r} = {2}{\cos}{\theta}[/latex] is

[latex]\hspace{3cm}\begin{align} V&=\underset{D}{\displaystyle\iint}f(r,\theta)r \ dr \ d\theta =\displaystyle\int_{\theta=0}^{\theta=\pi}\displaystyle\int_{r=0}^{r=2\cos\theta}(4-r^2)r \ dr \ d\theta \\ &=\displaystyle\int_{\theta=0}^{\theta=\pi}\left[4\frac{r^2}2-\frac{r^4}4\Bigg|_{r=0}^{r=2\cos\theta}\right]d\theta \\ &=\displaystyle\int_{0}^{\pi}[8\cos^2\theta-4\cos^2\theta]d\theta=\left[\frac52\theta+\frac52\sin\theta\cos\theta-\sin\theta\cos^3\theta\right]_0^{\pi}=\frac52\pi \end{align}[/latex].

First examine the region over which we need to set up the double integral and the accompanying paraboloid.

Figure 3. Finding the volume of a solid under a paraboloid and above a given triangle.

The region [latex]D[/latex] is [latex]{\left \{{(x,y)}{\mid}{0} \ {\leq} \ {x} \ {\leq} \ {1,x} \ {\leq} \ {y} \ {\leq} \ {2-x} \right \}}[/latex]. Converting the lines [latex]y=x[/latex], [latex]x=0[/latex], and [latex]x+y=2[/latex] in the [latex]xy[/latex]-plane to functions of [latex]r[/latex] and [latex]\theta[/latex], we have [latex]{{\theta} = {\pi}/{4}}, \ {{\theta} = {\pi}/{2}},[/latex] and [latex]{r} = {2}/{({cos}{\theta}+{sin}{\theta})}[/latex], respectively. Graphing the region on the [latex]xy[/latex]-plane, we see that it looks like [latex]{D}= {\left \{{(r,{\theta})}{\mid}{{\pi}/{4}} \ {\leq} \ {\theta} \ {\leq} \ {{\pi}/{2},0} \ {\leq} \ {r} \ {\leq} \ {2/{({\cos}{\theta}+{\sin}{\theta})}} \right \}}[/latex]. Now converting the equation of the surface gives [latex]{z} = {x^2} + {y^2} = {r^2}[/latex]. Therefore, the volume of the solid is given by the double integral
[latex]\hspace{2cm}\begin{align} V&=\underset{D}{\displaystyle\iint}f(r,\theta)r \ dr \ d\theta=\displaystyle\int_{\theta=\pi/4}^{\theta=\pi/2}\displaystyle\int_{r=0}^{r=2/(\cos\theta+\sin\theta}r^2r \ dr \ d\theta=\displaystyle\int_{\pi/4}^{\pi/2}\left[\frac{r^4}4\right]_{0}^{2/(\cos\theta+\sin\theta)} \ d\theta \\ &=\frac14\displaystyle\int_{\pi/4}^{\pi/2}\left(\frac2{(\cos\theta+\sin\theta)}\right)^4 \ d\theta = \frac{16}4\displaystyle\int_{\pi/4}^{\pi/2}\left(\frac1{(\cos\theta+\sin\theta)}\right)^4 \ d\theta =4\displaystyle\int_{\pi/4}^{\pi/2}\left(\frac1{(\cos\theta+\sin\theta)}\right)^4 \ d\theta. \end{align}[/latex]
As you can see, this integral is very complicated. So, we can instead evaluate this double integral in rectangular coordinates as

[latex]\large{V=\displaystyle\int_0^1\displaystyle\int_x^{2-x}(x^2+y^2)dydx}[/latex].

Evaluating gives
[latex]\hspace{6cm}\begin{align} V&=\displaystyle\int_0^1\displaystyle\int_x^{2-x}(x^2+y^2)dydx=\displaystyle\int_0^1\left[x^2y+\frac{y^3}3\right]\Bigg|_x^{2-x}dx \\ &=\displaystyle\int_0^1\frac83-4x+4x^2-\frac{8x^3}3dx \\ &=\left[\frac{8x}3-2x^2+\frac{4x^3}3-\frac{2x^4}3\right]\bigg|_0^1=\frac43. \end{align}[/latex]

Solution

The region [latex]D[/latex] for the integration is the base of the cone, which appears to be a circle on the [latex]xy[/latex]-plane (see the following figure).

Figure 4. Finding the volume of a solid inside the cone and above the [latex]xy[/latex]-plane.

We find the equation of the circle by setting [latex]z=0[/latex]:

[latex]\hspace{10cm}\large{\begin{align} {0} &= {2} - {\sqrt{{x^2}+{y^2}}} \\ {2} &= {\sqrt{{x^2}+{y^2}}} \\ {x^2} + {y^2} &= {4} \end{align}}[/latex]

This means the radius of the circle is 2, so for the integration we have [latex]{0} \ {\leq} \ {\theta} \ {\leq} \ {2{\pi}}[/latex] and [latex]{0} \ {\leq} \ {r} \ {\leq} \ {2}[/latex]. Substituting [latex]{x} = {r}{cos}{\theta}[/latex] and [latex]{y} = {r}{sin}{\theta}[/latex] in the equation [latex]{z} = {2} - {\sqrt{{x^2}+{y^2}}}[/latex] we have [latex]z=2-r[/latex]. Therefore, the volume of the cone is

[latex]\large{\displaystyle\int_{\theta=0}^{\theta=2\pi}\displaystyle\int_{r=0}^{r=2}(2-r)r \ dr \ d\theta = 2\pi\frac43=\frac{8\pi}3}[/latex] cubic units.

Analysis

Note that if we were to find the volume of an arbitrary cone with radius [latex]a[/latex] and height [latex]h[/latex] units, then the equation of the cone would be [latex]{z} = {h} - {\frac{h}{a}}{\sqrt{{x^2}+{y^2}}}[/latex].

We can still use Figure 4 and set up the integral as [latex]\displaystyle\int_{\theta=0}^{\theta=2\pi}\displaystyle\int_{r=0}^{r=a}\left(h-\frac{h}{a}\right)r \ dr \ d\theta[/latex].

Evaluating the integral, we get [latex]{\frac{1}{3}}{\pi}{a^2}{h}[/latex].

Sketching the graph of the function [latex]{r} = {\cos}{4}{\theta}[/latex] reveals that it is a polar rose with eight petals (see the following figure).

Figure 5. Finding the area of a polar rose with eight petals.

Using symmetry, we can see that we need to find the area of one petal and then multiply it by 8. Notice that the values of [latex]\theta[/latex] for which the graph passes through the origin are the zeros of the function [latex]{\cos}{4}{\theta}[/latex], and these are odd multiples of [latex]{\pi}/{8}[/latex]. Thus, one of the petals corresponds to the values of [latex]\theta[/latex] in the interval [latex][{{-\pi}/{8}},{{\pi}/{8}}][/latex]. Therefore, the area bounded by the curve [latex]{r} = {\cos}{4}{\theta}[/latex] is

A&=8\displaystyle\int_{\theta=-\pi/8}^{\theta=\pi/8}\displaystyle\int_{r=0}^{r=\cos4\theta}1 r \ dr \ d\theta \\ &=8\displaystyle\int_{-\pi/8}^{\pi/8}\left[\frac12r^2\bigg|_{0}^{\cos4\theta}\right]d\theta=8\displaystyle\int_{-\pi/8}^{\pi/8}\frac12\cos^2{4}\theta \ d\theta=8\left[\frac14\theta+\frac1{16}\sin4\theta\cos4\theta\bigg|_{-\pi/8}^{\pi/8}\right]=8[\frac{\pi}{16}]=\frac{\pi}2. \end{align}[/latex]

First and foremost, sketch the graphs of the region (Figure 6).

Figure 6. Finding the area enclosed by both a circle and a cardioid.

We can from see the symmetry of the graph that we need to find the points of intersection. Setting the two equations equal to each other gives

[latex]{3}{cos}{\theta} = {1} + {\cos}{\theta}[/latex].

One of the points of intersection is [latex]{\theta} = {{\pi}/{3}}[/latex]. The area above the polar axis consists of two parts, with one part defined by the cardioid from [latex]{\theta} = {0}[/latex] to [latex]{\theta} = {{\pi}/{3}}[/latex] and the other part defined by the circle from [latex]{\theta} = {{\pi}/{3}}[/latex] to [latex]{\theta} = {{\pi}/{2}}[/latex]. By symmetry, the total area is twice the area above the polar axis. Thus, we have

[latex]\large{A=2\left[\displaystyle\int_{\theta=0}^{\theta=\pi/3}\displaystyle\int_{r=0}^{r=1+\cos\theta}1r \ dr \ d\theta+\displaystyle\int_{\theta=\pi/3}^{\theta=\pi/2}\displaystyle\int_{r=0}^{r=3\cos\theta}1r \ dr \ d\theta\right]}[/latex].

Evaluating each piece separately, we find that the area is

[latex]{A} = {2}{\left ({{\frac{1}{4}}{\pi}} + {{\frac{9}{16}}{\sqrt{3}}} + {{\frac{3}{8}}{\pi}} - {{\frac{9}{16}}{\sqrt{3}}} \right)} = {2}{\left ({\frac{5}{8}}{\pi} \right )} = {{\frac{5}{4}}{\pi}}[/latex] square units.

This is an improper integral because we are integrating over an unbounded region [latex]R^2[/latex]. In polar coordinates, the entire plane [latex]R^2[/latex] can be seen as [latex]{0} \ {\leq} \ {\theta} \ {\leq} \ {{{2}{\pi}},0} \ {\leq} \ {r} \ {\leq} \ {\infty}[/latex].

Using the changes of variables from rectangular coordinates to polar coordinates, we have

[latex]\hspace{3cm}\begin{align} \underset{R^2}{\displaystyle\iint}e^{-10(x^2+y^2)}dx \ dy&=\displaystyle\int_{\theta=0}^{\theta=2\pi}\displaystyle\int_{r=0}^{r=\infty}e^{-10r^2}r \ dr \ d\theta=\displaystyle\int_{\theta=0}^{\theta=2\pi}\left(\displaystyle\lim_{a\to\infty}\displaystyle\int_{r=0}^{r=a}e^{-10r^2}r \ dr\right)d\theta \\ &=\left(\displaystyle\int_{\theta=0}^{\theta=2\pi}d\theta\right)\left(\displaystyle\lim_{a\to\infty}\displaystyle\int_{r=0}^{r=a}e^{-10r^2}r \ dr\right) \\ &=2\pi\left(\displaystyle\lim_{a\to\infty}\displaystyle\int_{r=0}^{r=a}e^{-10r^2}r \ dr\right) \\ &=2\pi\displaystyle\lim_{a\to\infty}\left(-\frac1{20}\right)\left(e^{-10r^2}\bigg|_0^1\right)\\ &=2\pi\left(-\frac1{20}\right)\displaystyle\lim_{a\to\infty}\left(e^{-10a^2}-1\right)\\ &=\frac{\pi}{10} \end{align}[/latex]