The bounds of the integral are determined solely by the region \(R\) over which we are integrating. In this case, it is a circle with equation \(x^2+y^2=1\text{.}\) We need to find polar bounds for this region. It may help to review Section 9.4; bounds for this circle are \(0\leq r\leq 1\) and \(0\leq \theta\leq 2\pi\text{.}\)

We replace \(f(x,y)\) with \(f(r\cos(\theta) ,r\sin(\theta) )\text{.}\) That means we make the following substitutions: \begin{equation*} 4-x-2y \Rightarrow 4-r\cos(\theta) -2r\sin(\theta) . \end{equation*}

Finally, we replace \(dA\) in the double integral with \(r\ dr\ d\theta\text{.}\) This gives the final iterated integral, which we evaluate: \begin{align*} \iint_Rf(x,y)\ dA \amp = \int_0^{2\pi}\int_0^1\big(4-r\cos(\theta) -2r\sin(\theta) \big)r\ dr\ d\theta\\ \amp = \int_0^{2\pi}\int_0^1\big(4r-r^2(\cos(\theta) -2\sin(\theta) )\big)\ dr\ d\theta\\ \amp = \int_0^{2\pi}\left.\left(2r^2-\frac13r^3(\cos(\theta) -2\sin(\theta) )\right)\right|_0^1d\theta\\ \amp = \int_0^{2\pi} \left(2-\frac13\big(\cos(\theta) -2\sin(\theta) \big)\right)\ d\theta\\ \amp = \left.\left(2\theta -\frac13\big(\sin(\theta) +2\cos(\theta) \big)\right)\right|_0^{2\pi}\\ \amp = 4\pi \approx 12.566. \end{align*}

The surface and region \(R\) are shown in Figure 13.3.6.

At first glance, this seems like a very hard volume to compute as the region \(R\) (shown in Figure 13.3.8(a)) has a hole in it, cutting out a strange portion of the surface, as shown in part (b) of the figure. However, by describing \(R\) in terms of polar equations, the volume is not very difficult to compute.

It is straightforward to show that the circle \((x-1)^2+y^2=1\) has polar equation \(r=2\cos(\theta)\text{,}\) and that the circle \((x-2)^2+y^2=4\) has polar equation \(r=4\cos(\theta)\text{.}\) Each of these circles is traced out on the interval \(0\leq\theta\leq\pi\text{.}\) The bounds on \(r\) are \(2\cos(\theta) \leq r\leq 4\cos(\theta)\text{.}\)

Replacing \(x\) with \(r\cos(\theta)\) in the integrand, along with replacing \(y\) with \(r\sin(\theta)\text{,}\) prepares us to evaluate the double integral \(\iint_Rf(x,y)\ dA\text{:}\) \begin{align*} \iint_Rf(x,y)\ dA \amp = \int_0^{\pi}\int_{2\cos(\theta) }^{4\cos(\theta) } \Big(4-\big(r\cos(\theta) -2\big)^2-\big(r\sin(\theta) \big)^2\Big)r\ dr\ d\theta\\ \amp = \int_0^{\pi}\int_{2\cos(\theta) }^{4\cos(\theta) } \big(-r^3+4r^2\cos(\theta) \big)\ dr\ d\theta\\ \amp = \int_0^\pi \left.\left(-\frac14r^4+\frac43r^3\cos(\theta) \right)\right|_{2\cos(\theta) }^{4\cos(\theta) }d\theta\\ \amp =\int_0^\pi \left(\left[-\frac14(256\cos^4(\theta) )+\frac43(64\cos^4(\theta) )\right]-\right.\\ \amp \ \left.\left[-\frac14(16\cos^4(\theta) )+\frac43(8\cos^4(\theta) )\right]\right)\ d\theta\\ \amp =\int_0^\pi\frac{44}3\cos^4(\theta) \ d\theta.\\ \end{align*} To integrate \(\cos^4(\theta)\text{,}\) rewrite it as \(\cos^2(\theta) \cos^2(\theta)\) and employ the power-reducing formula twice: \begin{align*} \cos^4(\theta) \amp =\cos^2(\theta) \cos^2(\theta) \\ \amp = \frac12\big(1+\cos(2\theta)\big)\frac12\big(1+\cos(2\theta)\big)\\ \amp = \frac14\big(1+2\cos(2\theta)+\cos^2(2\theta)\big)\\ \amp =\frac14\Big(1+2\cos(2\theta)+\frac12\big(1+\cos(4\theta)\big)\Big)\\ \amp = \frac38+\frac12\cos(2\theta)+\frac18\cos(4\theta).\\ \end{align*} Picking up from where we left off above, we have \begin{align*} \amp =\int_0^\pi\frac{44}3\cos^4(\theta) \ d\theta\\ \amp =\int_0^\pi \frac{44}3\left(\frac38+\frac12\cos(2\theta)+\frac18\cos(4\theta)\right)d\theta\\ \amp = \left.\frac{44}3\left(\frac{3}8\theta+\frac14\sin(2\theta)+\frac{1}{32}\sin(4\theta)\right)\right|_0^\pi\\ \amp =\frac{11}2\pi\approx 17.279. \end{align*}

While this example was not trivial, the double integral would have been much harder to evaluate had we used rectangular coordinates.

The region \(R\) we are integrating over is a circle with radius \(a\text{,}\) restricted to the first quadrant. Thus, in polar, the bounds on \(R\) are \(0\leq r\leq a\text{,}\) \(0\leq\theta\leq\pi/2\text{.}\) The integrand is rewritten in polar as \begin{equation*} \frac{1}{x^2+y^2+1} \Rightarrow \frac{1}{r^2\cos^2(\theta) +r^2\sin^2(\theta) +1} = \frac1{r^2+1}. \end{equation*}

We find the volume as follows: \begin{align*} \iint_Rf(x,y)\ dA \amp = \int_0^{\pi/2}\int_0^a\frac{r}{r^2+1}\ dr\ d\theta\\ \amp = \int_0^{\pi/2} \frac12\big(\ln\abs{r^2+1}\big)\Big|_0^a\ d\theta\\ \amp =\int_0^{\pi/2} \frac12\ln(a^2+1)\ d\theta\\ \amp = \left.\left(\frac12\ln(a^2+1)\theta\right)\right|_0^{\pi/2}\\ \amp = \frac{\pi}{4}\ln(a^2+1). \end{align*}

Figure 13.3.12 shows that \(f\) shrinks to near 0 very quickly. Regardless, as \(a\) grows, so does the volume, without bound.

The sphere of radius \(a\text{,}\) centered at the origin, has equation \(x^2+y^2+z^2=a^2\text{;}\) solving for \(z\text{,}\) we have \(z=\sqrt{a^2-x^2-y^2}\text{.}\) This gives the upper half of a sphere. We wish to find the volume under this top half, then double it to find the total volume.

The region we need to integrate over is the circle of radius \(a\text{,}\) centered at the origin. Polar bounds for this equation are \(0\leq r\leq a\text{,}\) \(0\leq\theta\leq2\pi\text{.}\)

All together, the volume of a sphere with radius \(a\) is: \begin{align*} 2\iint_R\sqrt{a^2-x^2-y^2}\ dA \amp = 2\int_0^{2\pi}\int_0^a\sqrt{a^2-(r\cos(\theta) )^2-(r\sin(\theta) )^2}r\ dr\ d\theta\\ \amp =2\int_0^{2\pi}\int_0^ar\sqrt{a^2-r^2}\ dr\ d\theta.\\ \end{align*} We can evaluate this inner integral with substitution. With \(u=a^2-r^2\text{,}\) \(du = -2r\ dr\text{.}\) The new bounds of integration are \(u(0) = a^2\) to \(u(a)=0\text{.}\) Thus we have: \begin{align*} \amp = \int_0^{2\pi}\int_{a^2}^0\big(-u^{1/2}\big)\ du\ d\theta\\ \amp = \int_0^{2\pi}\left.\left(-\frac23u^{3/2}\right)\right|_{a^2}^0 d\theta\\ \amp = \int_0^{2\pi}\left(\frac23a^3\right)\ d\theta\\ \amp = \left.\left(\frac23a^3\theta\right)\right|_0^{2\pi}\\ \amp = \frac43\pi a^3. \end{align*}

Generally, the formula for the volume of a sphere with radius \(r\) is given as \(4/3\pi r^3\text{;}\) we have justified this formula with our calculation.

From the outset, we should recognize that knowing how to set up this problem is probably more important than knowing how to compute the integrals. The iterated integral to come is not “hard” to evaluate, though it is long, requiring lots of algebra. Once the proper iterated integral is determined, one can use readily–available technology to help compute the final answer.

The region \(R\) that we are integrating over is bound by \(0\leq r\leq \cos(3\theta)\text{,}\) for \(0\leq \theta\leq\pi\) (note that this rose curve is traced out on the interval \([0,\pi]\text{,}\) not \([0,2\pi]\)). This gives us our bounds of integration. The integrand is \(z=1-x+0.1y\text{;}\) converting to polar, we have that the volume \(V\) is: \begin{equation*} V = \iint_R f(x,y)\ dA = \int_0^\pi\int_0^{\cos(3\theta)}\big(1-r\cos(\theta) +0.1r\sin(\theta) \big)r\ dr\ d\theta. \end{equation*}

Distributing the \(r\text{,}\) the inner integral is easy to evaluate, leading to \begin{equation*} \int_0^\pi \left(\frac12\cos^2(3\theta)-\frac13\cos^3(3\theta)\cos(\theta) +\frac{0.1}3\cos^3(3\theta)\sin(\theta) \right)\ d\theta. \end{equation*}

This integral takes time to compute by hand; it is rather long and cumbersome. The powers of cosine need to be reduced, and products like \(\cos(3\theta)\cos(\theta)\) need to be turned to sums using the Product To Sum formulas in the back cover of this text.

We rewrite \(\frac12\cos^2(3\theta)\) as \(\frac14(1+\cos(6\theta))\text{.}\) We can also rewrite \(\frac13\cos^3(3\theta)\cos(\theta)\) as: \begin{equation*} \frac13\cos^3(3\theta)\cos(\theta) = \frac13\cos^2(3\theta)\cos(3\theta)\cos(\theta) = \frac13\frac{1+\cos(6\theta)}2\big(\cos(4\theta)+\cos(2\theta)\big). \end{equation*}

This last expression still needs simplification, but eventually all terms can be reduced to the form \(a\cos(m\theta)\) or \(a\sin(m\theta)\) for various values of \(a\) and \(m\text{.}\)

We forgo the algebra and recommend the reader employ technology, such as WolframAlpha\textregistered, to compute the numeric answer. Such technology gives: \begin{equation*} \int_0^\pi\int_0^{\cos(3\theta)}\big(1-r\cos(\theta) +0.1r\sin(\theta) \big)r\ dr\ d\theta = \frac{\pi}{4} \approx 0.785u^3. \end{equation*}

Since the units were not specified, we leave the result as almost \(0.8\) cubic units (meters, feet, etc.) Should the artist want to scale the piece uniformly, so that each rose petal had a length other than 1, she should keep in mind that scaling by a factor of \(k\) scales the volume by a factor of \(k^3\text{.}\)