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It ain’t about how hard you can hit. Its about how hard you can get hit, and how much you can take, and keep moving forward, Rocky Balboa
Double integrals are indispensable mathematical tools. They allow us to calculate a variety of vital quantities across two-dimensional regions, such as surface areas, volumes, average values of functions, and physical properties like center of mass and moments of inertia.
While evaluating these integrals in Cartesian coordinates $(x, y)$ works well for rectangular regions, it can become a mathematical nightmare when dealing with curves. This is where polar coordinates come to the rescue.
Double integrals are a powerful tool in calculus for finding areas, volumes, and other quantities over a two-dimensional region. Sometimes, evaluating these integrals in Cartesian coordinates (using x and y) can be complex or cumbersome. In such cases, converting to polar coordinates can simplify the problem.
Polar coordinates are particularly useful when dealing with regions that have circular symmetry, such as disks, annuli (rings), or sectors Instead of describing a point in the plane using Cartesian coordinates $(x, y)$, polar coordinates describe a point using:
Basically, in polar coordinates, we replace x and y with $r\cos \theta$ and $r \sin \theta$, respectively, and adjust the limits of integration accordingly.
Consider the integral $\int \int_{x^2+y^2 \leq 1, x, y \ge 0} (1-x^2-y^2)dA$ over the region, $x^2 + y^2 \leq 1; x, y \ge 0$. Here, R is the region in the first quadrant bounded by the unit circle.
To convert from Cartesian coordinates $(x, y)$ to polar coordinates $(r, \theta)$, use the following relationships: $x = r\cos(\theta), y = r\sin(\theta)$.
We fix $\theta$, $r$ varies from 0 to 1. Then, $\theta$ ranges from $0$ to $\frac{\pi}{2}$.
When converting a double integral, we cannot simply swap $dx \, dy$ for $dr \, d\theta$. We must account for how area scales in the new coordinate system.
Imagine a tiny “polar rectangle” defined by a small change in radius $\Delta r$ and a small change in angle $\Delta \theta$. Unlike a Cartesian grid where every square is the exact same size, the size of a polar rectangle depends on how far it is from the origin.
Recall that the sine function $\sin(\Delta \theta)$ relates the angle ($\Delta \theta$) to the ratio of the length of the side opposite the angle to the length of the hypotenuse in a right triangle, and when $\Delta \theta$ is very close to 0, we can make an approximation: $\sin(\Delta \theta) \approx \Delta \theta$, hence $\Delta A \approx \Delta r \cdot r\sin(\Delta \theta) \approx \Delta r \cdot r\Delta \theta$. In the infinitesimal scale, $dA = dr \cdot rd\theta, \int \int fdA = \int (\int f\cdot rdr)d\theta$.
Therefore, our general formula for conversion becomes: $\iint_R f(x,y) \, dx \, dy = \iint_S f(r\cos\theta, r\sin\theta) \, r \, dr \, d\theta$
Let’s find the volume under the surface $z = 1 - x^2 - y^2$ restricted to the first quadrant of the unit circle ($x^2 + y^2 \le 1$ where $x \ge 0, y \ge 0$).
Set up the conversion.
Substitute $x^2 + y^2 = r^2$. The function becomes $f(r, \theta) = 1 - r^2$.
Limits of Integration.
Because the region is the unit circle, the radius $r$ extends from the origin out to the boundary edge: $0 \le r \le 1$. Because we are restricted to the first quadrant, the angle sweeps from the $x$-axis to the $y$-axis: $0 \le \theta \le \frac{\pi}{2}$.
Set Up the Integral.
$\int \int_{x^2+y^2 \leq 1, x, y \ge 0} (1-x^2-y^2)dx\,dy = \int_{0}^{\frac{\pi}{2}} (\int_{0}^{1} (1-r^2)rdr)d\theta$.
Compute the Inner Integral.
We first evaluate the integral with respect to r, keeping $\theta$ fixed: $\int_{0}^{1} (1-r^2)rdr = \frac{r^2}{2}-\frac{r^4}{4}\bigg|_{0}^{1} = \frac{1}{2}-\frac{1}{4} = \frac{1}{4}$
Compute the Outer Integral.
Next, we integrate with respect to $\theta$: $\int_{0}^{\frac{\pi}{2}} \frac{1}{4}d\theta = \frac{1}{4}\theta\bigg|_{0}^{\frac{\pi}{2}} = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8}$.
Conclusion. The volume under the surface $z = 1 −x^2 -y^2$ over the region in the first quadrant bounded by the unit circle is: $\frac{\pi}{8}$.
Here’s a step-by-step approach for converting and evaluating integrals in polar coordinates:
Identify the surfaces and the region
$z^2 - x^2 - y^2 = 1$ matches the standard form equation $\frac{z^2}{c^2} - \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (where $a=1$, $b=1$, and $c=1$), which defines a hyperboloid of two sheets opening along the z-axis.
The lower surface can be rewritten as $z^2 = 1 + x^2 + y^2$. Because we are looking at the volume bounded above by $z=2$, we take the positive root: $z = \sqrt{1+x^2+y^2}$. Geometrically, this represents the upper sheet of a hyperboloid of two sheets.
We set the two equations equal to each other because the boundary of the solid is formed exactly where the two surfaces intersect.
Think of it like a bowl (the hyperboloid) covered by a flat lid (the plane $z=2$). The widest part of the solid —which defines its “shadow” or footprint on the xy-plane—occurs precisely where the bowl meets the lid. At that intersection rim, the $z$-value of the bowl must equal the $z$-value of the flat plane: $2 = \sqrt{1+x^2+y^2}$
$4 = 1 + x^2 + y^2 \implies x^2 + y^2 = 3$
The region of integration $R$ is a disk centered at the origin with a radius of $\sqrt{3}$.
Convert to Polar
The bounds of the disk are $0 \le r \le \sqrt{3}$ and $0 \le \theta \le 2\pi$.
The height of the solid at any point is the upper surface minus the lower surface: $z_{\text{top}} - z_{\text{bottom}} = 2 - \sqrt{1+r^2}$.
Set up the double integral.
Therefore, the volume V is given by the double integral: $V = \int_{0}^{2\pi} \int_{0}^{\sqrt{3}} (2-\sqrt{1+r^2})rdrd\theta$ where R is the disk, $\theta$ ranges from 0 to $2\pi$ and $0 \leq r \leq \sqrt{3}$.
Evaluate the radial integral.
We first integrate with respect to r:
$\int_{0}^{\sqrt{3}} (2-\sqrt{1+r^2})rdr =[\text{Distribute the r:}] \int_{0}^{\sqrt{3}} (2r-r\sqrt{1+r^2})dr$. Use a simple u-substitution for the second term: $u = 1+ r^2, du = 2rdr$.
$\int_{0}^{\sqrt{3}} 2rdr -\int_{0}^{\sqrt{3}}r\sqrt{u}\frac{du}{2r} = \int_{0}^{\sqrt{3}} 2rdr -\int_{0}^{\sqrt{3}}u^{\frac{1}{2}}\frac{du}{2} = r^2-\frac{1}{3}u^{\frac{3}{2}} = r^2-\frac{1}{3}(1+r^2) ^{\frac{3}{2}}\bigg|_{0}^{\sqrt{3}}$
$3-\frac{1}{3}4^{\frac{3}{2}}+\frac{1}{3} = 3-\frac{1}{3}\sqrt{4}^{3}+\frac{1}{3} = 3-\frac{8}{3}+\frac{1}{3} = 3-\frac{7}{3} = \frac{9}{3}-\frac{7}{3} = \frac{2}{3}$
Evaluate the angular integral: $V = \int_{0}^{2\pi} \frac{2}{3}d\theta = \frac{2}{3}\theta\bigg|_{0}^{2\pi} = \frac{4\pi}{3}.$
Identify the Region of Integration.
$\star$ $x^2 + y^2 \ge 1$ means we are outside a circle of radius 1.
$\star$ $x^2 + y^2 \le 4$ means we are inside a circle of radius 2.
$\star$ $x \le 0$ restricts our region to the left half of the $xy$-plane (Quadrants II and III).
This forms a semi-annulus (a ring-shaped region) between the circles of radii 1 and 2, lying in the left half-plane. Refer to Figure A for a visual representation and aid in understanding it.
Setup Limits and Convert
The radius bounds are directly given: $1 \le r \le 2$.
Because we are strictly in the left half-plane, our angle sweeps from the positive $y$-axis to the negative $y$-axis: $\frac{\pi}{2} \le \theta \le \frac{3\pi}{2}$.
The function $(x+y)$ becomes $(r\cos\theta + r\sin\theta)$.
Set up the double integral. $I = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \int_{1}^{2} (r\cos\theta + r\sin\theta) \, r \, dr \, d\theta$
Evaluate the radial integral.
$\int_{1}^{2} (r\cos(\theta)+r\sin(\theta))rdr = (\cos(\theta)+\sin(\theta))\int_{1}^{2} r^2dr = (\cos(\theta)+\sin(\theta)) \cdot \frac{r^3}{3}\bigg|_{1}^{2} = (\cos(\theta)+\sin(\theta))·(\frac{8}{3}-\frac{1}{3}) = (\cos(\theta)+\sin(\theta)) \cdot \frac{7}{3}$
Evaluate the angular integral.
$\iint_R (x + y) dA = \frac{7}{3} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (\cos(\theta)+\sin(\theta)) \cdot d\theta = \frac{7}{3} (\sin(\theta)-\cos(\theta))\bigg|_{\frac{\pi}{2}}^{\frac{3\pi}{2}}$
$\frac{7}{3} (\sin(\frac{3\pi}{2})-cos(\frac{3\pi}{2})-(sin(\frac{\pi}{2})-cos(\frac{\pi}{2}))) = \frac{7}{3}(-1-0-(1-0)) = \frac{-14}{3}$
Geometric Insight: Why is the result negative? The function $f(x,y) = x+y$ produces $z$-values. Because our region lies entirely where $x$ is negative, and spans symmetrically across the $x$-axis (meaning the positive $y$ and negative $y$ values cancel each other out), the net “volume” dips below the $xy$-plane.
Determine Intersection.
Look at the standard parent paraboloid: $z = x^2 + y^2$. As x and y get larger (positive or negative), z gets larger and more positive. This creates a standard bowl opening upward.
When you multiply by a negative number to get $z = -3x^2 - 3y^2$, the math flips. Because any real number squared is positive, $x^2$ and $y^2$ are always positive. Multiplying them by -3 forces the result to always be negative.
As you move away from the center $(0, 0)$, z drops lower and lower into negative numbers. This creates the upside-down bowl (downward-opening) shape.
If you look at the exact center where x = 0 and y = 0: $z = 10 - 3(0)^2 - 3(0)^2 = 10$. This point $(0, 0, 10)$ is the highest point (the peak or vertex) of the paraboloid.
To sum up, $-3x^2 - 3y^2$ is an upside-down (downward opening) paraboloid, shifted up 10 units. It is sliced horizontally by the plane $z = 4$.
To find their intersection: $10 - 3x^2 - 3y^2 = 4 \implies 3x^2 + 3y^2 = 6 \implies x^2 + y^2 = 2$
Convert to Polar Coordinates.
In polar coordinates, $x = r\cos(\theta)$ and $y = r\sin(\theta)$.
The shadow $R$ is a circle of radius $\sqrt{2}$. So, $0 \le r \le \sqrt{2}$ and $0 \le \theta \le 2\pi$ (Figure B).
Express the Height Function.
The height of the solid above the xy-plane ($z = 4$) at any point $(r, \theta)$ is given by the difference between the paraboloid and the plane: $(z_{\text{top}} - z_{\text{bottom}}) = (10 - 3x^2 - 3y^2) - (4) = 6 - 3(x^2 + y^2) = 6 - 3r^2$
Set Up the Double Integral.
$V = \int_{0}^{2\pi}(\int_{0}^{\sqrt{2}} (6-3r^2)rdr)d\theta$
Evaluate the Radial Integral.
$\int_{0}^{\sqrt{2}} (6-3r^2)rdr = \int_{0}^{\sqrt{2}} (6r-3r^3)dr = 3r^2-\frac{3}{4}r^4\bigg|_{0}^{\sqrt{2}} = 3\cdot 2-\frac{3}{4}·4 = 6 - 3 = 3.$
Evaluate the Angular Integral.
$V = \int_{0}^{2\pi}(\int_{0}^{\sqrt{2}} (6-3r^2)rdr)d\theta = \int_{0}^{2\pi} 3d\theta = 3\theta\bigg|_{0}^{2\pi} = 6\pi.$
Conclusion: The volume of the solid bounded by the paraboloid $10 - 3x^2 - 3y^2$ and the plane $z = 4$ is $6\pi$.
Determine the Region of intersection.
To find the limits of our double integral, we first need to know where these two surfaces intersect. The intersection forms the widest part of our solid, and its projection onto the $xy$-plane will be our region of integration, $D$.
Set the two $z$-equations equal to each other: $3x^2 + 3y^2 = 4 - x^2 - y^2$
Combine like terms by moving all $x$ and $y$ variables to the left side: $4x^2 + 4y^2 = 4$
Divide by 4: $x^2 + y^2 = 1$
This intersection projects a perfect circle of radius $r = 1$ centered at the origin in the $xy$-plane.
The z-value in this region is $z = 3x^2 + 3y^2= 3(x^2 + y^2) = 3 \cdot 1 = 3$ (Figure C).
Convert to Polar Coordinates.
Because our region of integration $D$ is a circle, polar coordinates are the perfect tool. Using the standard conversions $x = r\cos(\theta)$, $y = r\sin(\theta)$, and $x^2 + y^2 = r^2$, the region $D$ becomes incredibly simple to define: $0 \le r \le 1$, $0 \le \theta \le 2\pi$.
Express the height function.
The volume of a solid is the double integral of its height. The height of this solid at any given point $(r, \theta)$ is the “roof” surface minus the “floor” surface.
$\text{Height} = (4 - x^2 - y^2) - (3x^2 + 3y^2) = 4 - 4x^2 - 4y^2 = 4 - 4(x^2 + y^2)$
Substituting our polar identity $x^2 + y^2 = r^2$: $\text{Height} = 4(1 - r^2)$
Set Up the Double Integral.
Recall that when converting a double integral to polar coordinates, the area differential $dA$ becomes $r \, dr \, d\theta$. Do not forget the extra $r$!
The volume V of the solid can be found by integrating the height over the region: $V = \iint_{D} \text{Height} \, dA = \int_{0}^{2\pi} \int_{0}^{1} 4(1 - r^2)r \, dr \, d\theta$
Evaluate the Radial Integral.
First, evaluate the inner radial integral. Distribute the $r$ to make integration simple: $4 \int_{0}^{1} (r - r^3) \, dr = 4 \left( \frac{r^2}{2} - \frac{r^4}{4} \right) \bigg\vert{}_{0}^{1}$
$= 4 \left( \frac{1}{2} - \frac{1}{4} \right) = 4 \left( \frac{1}{4} \right) = 1$
Now, evaluate the outer angular integral: $V = \int_{0}^{2\pi} 1 \, d\theta = \theta \bigg\vert{}_{0}^{2\pi} = 2\pi$
Conclusion. The volume of the solid is $2\pi$ cubic units.
Understand the Geometric Boundaries.
The three surfaces intersect as follows:
$\star$ Sphere: $x^2 + y^2 + z^2 = 9$ is a sphere of radius 3 centered at the origin.
$\star$ Cylinder: $x^2 + y^2 = 5$ is a cylinder of radius $\sqrt{5}$ extending infinitely along the $z$-axis.
$\star$ Plane: $z = 0$ is the flat $xy$-plane (the floor).
We are essentially using the cylinder like an apple corer, cutting straight down through the sphere. We want the volume of that core from the $xy$-plane up to the curved top of the sphere.
Because the cylinder strictly restricts our boundaries on the $xy$-plane, our region of integration $D$ is just the disk $x^2 + y^2 \le 5$.
Furthermore, the region that we want the volume for is really a cylinder with a cap that comes from the sphere: $x^2 + y^2 + z^2 = 9$. Since the cylinder is $x^2 + y^2 = 5$, our cap is $5 + z^2 = 9 \implies z^2 = 4 \implies z = 2$ (Figure D).
Convert to Polar Coordinates.
Our region $D$ is a disk of radius $\sqrt{5}$. In polar coordinates, this is simply bounded by: $0 \le r \le \sqrt{5}$, $0 \le \theta \le 2\pi$.
Express the height function.
Our “roof” is the upper hemisphere.
The height z is bounded below by the plane $z = 0$ and above by the sphere. From the sphere’s equation: $z = \sqrt{9-x^2-y^2}= \sqrt{9-r^2}$
Set Up the Double Integral.
The volume $V$ is the double integral of the height function $z = \sqrt{9-r^2}$ over the region $D$: $V = \iint_{D} \sqrt{9 - x^2 - y^2} \, dA = \int_{0}^{2\pi} \int_{0}^{\sqrt{5}} \sqrt{9 - r^2} \, r \, dr \, d\theta$
Evaluate the Radial Integral.
To evaluate the inner integral, we use $u$-substitution:
Let $u = 9 - r^2$. Then $du = -2r \, dr$, which means $r \, dr = -\frac{1}{2}du$.
Next, change the limits of integration:
$\star$ When $r = 0$, $u = 9$.
$\star$ When $r = \sqrt{5}$, $u = 9 - 5 = 4$.
Now, rewrite the radial integral in terms of $u$: $\int_{0}^{\sqrt{5}} r(9 - r^2)^{\frac{1}{2}} \, dr = \int_{9}^{4} \left(-\frac{1}{2}\right)u^{\frac{1}{2}} \, du$
We can use the negative sign to flip the limits of integration back to a standard low-to-high order: $= \frac{1}{2} \int_{4}^{9} u^{\frac{1}{2}} \, du = \frac{1}{2} \left[ \frac{2}{3}u^{\frac{3}{2}} \right]_{4}^{9}$
$= \frac{1}{3} \left( 9^{\frac{3}{2}} - 4^{\frac{3}{2}} \right)$
Since $9^{\frac{3}{2}} = (\sqrt{9})^3 = 3^3 = 27$ and $4^{\frac{3}{2}} = (\sqrt{4})^3 = 2^3 = 8$: $= \frac{1}{3} (27 - 8) = \frac{19}{3}$
Finally, evaluate the angular integral: $V = \int_{0}^{2\pi} \frac{19}{3} \, d\theta = \frac{19}{3}\theta \bigg\vert{}_{0}^{2\pi} = \frac{38\pi}{3}$
Understand the Geometric Boundaries.
The curve $r = \cos(4\theta)$ is a polar equation that represents a type of graph known as a “polar rose”. These are curves that have petal-like structures. The number of petals is determined by the coefficient of $\theta$, k, inside the cosine function.
If $k$ is odd, there are $k$ petals. If $k$ is even, there are $2k$ petals. Here, $k=4$, resulting in an 8-petaled rose.
Refer to Figure 1 for a visual representation and aid in understanding it.
Exploit Symmetry to Find the Bounds.
The petals are symmetrically distributed around the origin. Since the cosine function is periodic with period $2\pi$, it repeats its values.
This means the area of all petals is the same. Therefore, instead of calculating the area of the entire curve directly, we can calculate the area of one petal and then multiply by the number of petals (which is 8 in this case) to get the total area.
Where does one petal begin and end? A petal is formed when the radius $r$ starts at $0$, grows to its maximum length, and returns to $0$. We find these boundaries by setting $r=0$: $\cos(4\theta) = 0$
The cosine function equals zero at odd multiples of $\pi/2$. Therefore: $4\theta = -\frac{\pi}{2} \quad \text{and} \quad 4\theta = \frac{\pi}{2}$
$\theta = -\frac{\pi}{8} \quad \text{and} \quad \theta = \frac{\pi}{8}$
Thus, one petal is traced out as $\theta$ ranges from $-\pi/8$ to $\pi/8$.
Set Up the Double Integral.
To find a flat 2D area using double integrals, we integrate the constant function $f(r,\theta) = 1$ over the region.
$\text{Total Area} = 8 \times (\text{Area of one petal})$
$\text{Area} = 8 \int_{-\pi/8}^{\pi/8} \int_{0}^{\cos(4\theta)} 1 \, r \, dr \, d\theta$
Evaluate the integral.
Evaluate the inner integral, $\int_{0}^{\cos(4\theta)} r \, dr = \frac{r^2}{2} \bigg\vert{}_{0}^{\cos(4\theta)} = \frac{\cos^2(4\theta)}{2}$
Substitute this back into the outer integral: $\text{Area} = 8 \int_{-\pi/8}^{\pi/8} \frac{\cos^2(4\theta)}{2} \, d\theta = 4 \int_{-\pi/8}^{\pi/8} \cos^2(4\theta) \, d\theta$
To integrate a squared cosine function, we must use the power-reduction identity: $\cos^2(x) = \frac{1 + \cos(2x)}{2}$. By substituting $x = 4\theta$, the identity becomes $\cos^2(4\theta) = \frac{1 + \cos(8\theta)}{2}$.
$\text{Area} = 4 \int_{-\pi/8}^{\pi/8} \frac{1 + \cos(8\theta)}{2} \, d\theta = 2 \int_{-\pi/8}^{\pi/8} (1 + \cos(8\theta)) \, d\theta$
Now, we just integrate directly: $\text{Area} = 2 \left[ \theta + \frac{1}{8}\sin(8\theta) \right]_{-\pi/8}^{\pi/8}$
$= 2 \left[ \left( \frac{\pi}{8} + \frac{1}{8}\sin(\pi) \right) - \left( -\frac{\pi}{8} + \frac{1}{8}\sin(-\pi) \right) \right]$
Since $\sin(\pi) = 0$ and $\sin(-\pi) = 0$: $= 2 \left[ \frac{\pi}{8} - \left(-\frac{\pi}{8}\right) \right] = 2 \left[ \frac{2\pi}{8} \right] = 2 \left[ \frac{\pi}{4} \right] = \frac{\pi}{2}$
Conclusion. The total area of the 8-petaled rose is $\pi/2$ square units.
Understand the Geometric Boundaries.
The equation $r = 1−\sin(\theta)$ describes a cardioid (Refer to Figure 2 for a visual representation and aid in understanding it), which is a heart-shaped curve. This cardioid is symmetric about the vertical axis (the line $\theta = \pi/2$).
Let’s understand what we are calculating. The height function $f(r, \theta) = r$ in rectangular coordinates is $z = \sqrt{x^2+y^2}$, which mathematically describes an infinite cone stretching upwards. We are essentially finding the volume of this cone, but we are using a heart-shaped cardioid cookie cutter as the base on the $xy$-plane.
Set Up the Double Integral.
The volume element $dV$ at any point $(r, \theta)$ is the height multiplied by the area element, which in polar space is $(r)(r \, dr \, d\theta) = r^2 \, dr \, d\theta$.
The cardioid traces out a complete shape over one full rotation, so $\theta$ ranges from $0$ to $2\pi$.
$V = \int_{0}^{2\pi} \int_{0}^{1-\sin(\theta)} r \cdot r \, dr \, d\theta = \int_{0}^{2\pi} \int_{0}^{1-\sin(\theta)} r^2 \, dr \, d\theta$
Evaluate the Inner Radial Integral.
$\int_{0}^{1-\sin(\theta)} r^2 \, dr = \frac{r^3}{3} \bigg\vert{}_{0}^{1-\sin(\theta)} = \frac{(1-\sin(\theta))^3}{3}$
Evaluate the Outer Angular Integral.
Now substitute this back into the volume formula and factor out the constant: $V = \frac{1}{3} \int_{0}^{2\pi} (1-\sin(\theta))^3 \, d\theta$
We need to expand the cubic binomial $(A-B)^3 = A^3 - 3A^2B + 3AB^2 - B^3$: $V = \frac{1}{3} \int_{0}^{2\pi} (1 - 3\sin(\theta) + 3\sin^2(\theta) - \sin^3(\theta)) \, d\theta$
We can evaluate this by splitting it into four separate integrals over the period $[0, 2\pi]$:
$\star (i)$ $\int_{0}^{2\pi} 1 \, d\theta = 2\pi$
$\star (ii)$ $-3 \int_{0}^{2\pi} \sin(\theta) \, d\theta = 0$
$\star (iii)$ $3 \int_{0}^{2\pi} \sin^2(\theta) \, d\theta$
Using the half-angle identity $\sin^2(\theta) = \frac{1-\cos(2\theta)}{2}$: $= 3 \int_{0}^{2\pi} \frac{1 - \cos(2\theta)}{2} \, d\theta = \frac{3}{2} \left[ \theta - \frac{1}{2}\sin(2\theta) \right]_{0}^{2\pi} = \frac{3}{2}(2\pi - 0) = 3\pi$
$\star (iv)$ $-\int_{0}^{2\pi} \sin^3(\theta) \, d\theta = 0$
Why do the integrals of $\sin(\theta)$ and $\sin^3(\theta)$ evaluate to zero over the interval $[0, 2\pi]$? It has to do with the periodicity and shift symmetry of the sine wave. For every positive y-value sine takes between $0$ and $\pi$, it takes the exact corresponding negative y-value between $\pi$ and $2\pi$ (mathematically, $\sin(\theta+\pi) = -\sin(\theta)$). Because the area above the $x$-axis perfectly mirrors the area below the $x$-axis, they sum to zero. This holds true for any odd power of $\sin(\theta)$ or $\cos(\theta)$ integrated over a full period.
Finally, substitute these four evaluated components back into the volume expression: $V = \frac{1}{3} [2\pi - 0 + 3\pi - 0]$
$V = \frac{1}{3} (5\pi) = \frac{5\pi}{3}$
Conclusion. The volume of the solid is $5\pi/3$ cubic units.
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