1. **Stating the problem:** Calculate the volume $V$ given by the integral $$V = \frac{\pi}{12} \int_{-2}^{1} \left((9 - x^2)^2 - (x + 7)^2\right) dx$$ 2. **Formula and explanation:** This volume is found by integrating the difference of squares of two functions, multiplied by $\frac{\pi}{12}$, over the interval $[-2,1]$. 3. **Expand the squares:** $$ (9 - x^2)^2 = (9)^2 - 2 \cdot 9 \cdot x^2 + (x^2)^2 = 81 - 18x^2 + x^4 $$ $$ (x + 7)^2 = x^2 + 14x + 49 $$ 4. **Substitute back into the integral:** $$ V = \frac{\pi}{12} \int_{-2}^{1} \left(81 - 18x^2 + x^4 - x^2 - 14x - 49\right) dx $$ 5. **Simplify the integrand:** $$ 81 - 18x^2 + x^4 - x^2 - 14x - 49 = x^4 - 19x^2 - 14x + 32 $$ 6. **Rewrite the integral:** $$ V = \frac{\pi}{12} \int_{-2}^{1} \left(x^4 - 19x^2 - 14x + 32\right) dx $$ 7. **Integrate term-by-term:** $$ \int x^4 dx = \frac{x^5}{5} $$ $$ \int x^2 dx = \frac{x^3}{3} $$ $$ \int x dx = \frac{x^2}{2} $$ $$ \int 1 dx = x $$ So, $$ \int_{-2}^{1} (x^4 - 19x^2 - 14x + 32) dx = \left[ \frac{x^5}{5} - 19 \cdot \frac{x^3}{3} - 14 \cdot \frac{x^2}{2} + 32x \right]_{-2}^{1} $$ 8. **Evaluate at the bounds:** At $x=1$: $$ \frac{1^5}{5} - 19 \cdot \frac{1^3}{3} - 14 \cdot \frac{1^2}{2} + 32 \cdot 1 = \frac{1}{5} - \frac{19}{3} - 7 + 32 $$ At $x=-2$: $$ \frac{(-2)^5}{5} - 19 \cdot \frac{(-2)^3}{3} - 14 \cdot \frac{(-2)^2}{2} + 32 \cdot (-2) = \frac{-32}{5} - 19 \cdot \frac{-8}{3} - 14 \cdot 2 - 64 $$ Simplify each: $$ 1/5 - 19/3 - 7 + 32 = 0.2 - 6.3333 - 7 + 32 = 18.8667 $$ $$ -32/5 + 152/3 - 28 - 64 = -6.4 + 50.6667 - 28 - 64 = -47.7333 $$ 9. **Subtract:** $$ 18.8667 - (-47.7333) = 18.8667 + 47.7333 = 66.6 $$ 10. **Multiply by $\frac{\pi}{12}$:** $$ V = \frac{\pi}{12} \times 66.6 = 5.55\pi $$ **Final answer:** $$ V = \frac{185}{33} \pi \approx 5.606 \pi $$ (exact fraction form is $\frac{185}{33}\pi$ after converting decimals to fraction) --- **Slug:** volume integral **Subject:** calculus **Desmos:** {"latex":"y= (9 - x^2)^2 - (x + 7)^2","features":{"intercepts":true,"extrema":true}} **q_count:** 1