Question Statement

Find the volume of the solid that results when the shaded region is revolved about the indicated axis:

(i) $y=\sqrt{3-x}$ about the $x$-axis.

(ii) $y=3-2x$ about the $y$-axis.

---

Background and Explanation

These problems use the disk method for finding volumes of solids of revolution. When revolving a region bounded by $y=f(x)$ about the $x$-axis (or $x=g(y)$ about the $y$-axis), we integrate the cross-sectional area of circular disks along the axis of rotation. The radius of each disk is given by the function value at that point.

---

Solution

Part (i): $y=\sqrt{3-x}$ about the $x$-axis

We use the disk method formula for rotation about the $x$-axis: $V=\pi {\int }_a^b[f(x){]}^{2}dx$

From the problem and graph, we identify:

• The function: $f(x)=\sqrt{3-x}$
• The limits of integration: $a=-1$ and $b=3$

Substituting into the volume formula: $V=\pi {\int }_{-1}^3(\sqrt{3-x}{)}^{2}dx$

Simplifying the integrand by squaring the square root: $V=\pi {\int }_{-1}^3(3-x)dx$

We can split this into two separate integrals for easier evaluation: $V=\pi \left[3{\int }_{-1}^{3}1dx-{\int }_{-1}^{3}xdx\right]$

Finding the antiderivatives: $V=\pi \left[3{\left[x\right]}_{-1}^3-{\left[\frac{x^2}{2}\right]}_{-1}^3\right]$

Applying the limits of integration: $V=\pi \left[3(3-(-1))-\frac{1}{2}\left((3{)}^2-(-1{)}^2\right)\right]$

Simplifying the arithmetic inside the brackets: $V=\pi \left[3(4)-\frac{1}{2}(9-1)\right]$ $V=\pi \left[12-\frac{1}{2}(8)\right]$ $V=\pi [12-4]$ $V=8\pi$

Part (ii): $y=3-2x$ about the $y$-axis

For rotation about the $y$-axis, we must express $x$ as a function of $y$ and integrate with respect to $y$: $V=\pi {\int }_c^d[g(y){]}^{2}dy$

First, solve the equation $y=3-2x$ for $x$: $2x=3-y$ $x=\frac{3-y}{2}$

From the graph, the limits for $y$ are from $0$ to $3$, so $c=0$ and $d=3$. Thus $g(y)=\frac{3-y}{2}$.

Setting up the volume integral: $V=\pi {\int }_0^3{\left(\frac{3-y}{2}\right)}^{2}dy$

Expanding the squared term in the integrand: $V=\pi {\int }_0^3\frac{(3-y{)}^2}{4}dy=\frac{\pi }{4}{\int }_0^3(9-6y+y^2)dy$

Separating into individual integrals: $V=\frac{\pi }{4}\left[9{\int }_0^{3}1dy+{\int }_0^3{y}^{2}dy-6{\int }_0^{3}ydy\right]$

Evaluating each antiderivative: $V=\frac{\pi }{4}\left[9{\left[y\right]}_0^3+{\left[\frac{y^3}{3}\right]}_0^3-6{\left[\frac{y^2}{2}\right]}_0^3\right]$

Simplifying the third term ($6\cdot \frac{y^2}{2}=3y^2$) and applying limits: $V=\frac{\pi }{4}\left[9(3-0)+\frac{1}{3}(3^3-0^3)-3(3^2-0^2)\right]$

Calculating each term: $V=\frac{\pi }{4}\left[27+\frac{1}{3}(27)-3(9)\right]$ $V=\frac{\pi }{4}\left[27+9-27\right]$ $V=\frac{\pi }{4}\cdot 9$ $V=\frac{9\pi }{4}$

---

Key Formulas or Methods Used

• Disk Method about $x$-axis: $V=\pi {\int }_a^b[f(x){]}^{2}dx$
• Disk Method about $y$-axis: $V=\pi {\int }_c^d[g(y){]}^{2}dy$ where $x=g(y)$
• Power Rule for Integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ (for $n\ne -1$)
• Algebraic expansion: $(3-y{)}^2=9-6y+y^2$

---

Summary of Steps

1. Identify the axis of rotation to determine whether to integrate with respect to $x$ or $y$
2. Set up the appropriate disk method formula based on the axis of rotation
3. Determine the limits of integration from the given region or graph (check where the curve intersects the boundaries)
4. Express the radius function as $f(x)$ if rotating about the $x$-axis, or solve for $x=g(y)$ if rotating about the $y$-axis
5. Square the radius function and set up the definite integral
6. Expand and separate the integral into simpler terms if necessary
7. Evaluate using antiderivatives and apply the Fundamental Theorem of Calculus
8. Simplify the arithmetic to obtain the final volume answer