Let’s solve each problem step by step: --- ### **5)** **Given:** - \( y = 7x^2 \) and \( y = 72 - x^2 \) - Revolve the region bounded by these curves about the x-axis. #### **Step 1: Find the points of intersection** Set \( 7x^2 = 72 - x^2 \): \[ 7x^2 + x^2 = 72 \implies 8x^2 = 72 \implies x^2 = 9 \implies x = \pm 3 \] #### **Step 2: Set up the volume integral (Washer Method)** The outer radius \( R_{out} = y_{top} = 72 - x^2 \) The inner radius \( R_{in} = y_{bottom} = 7x^2 \) \[ V = \pi \int_{-3}^{3} \left[(72 - x^2)^2 - (7x^2)^2\right] dx \] #### **Step 3: Simplify the integrand** \[ (72 - x^2)^2 = 5184 - 144x^2 + x^4 \\ (7x^2)^2 = 49x^4 \] So, \[ (72 - x^2)^2 - (7x^2)^2 = 5184 - 144x^2 + x^4 - 49x^4 = 5184 - 144x^2 - 48x^4 \] #### **Step 4: Integral calculation** \[ V = \pi \int_{-3}^{3} (5184 - 144x^2 - 48x^4) dx \] Because the function is even, we can double the integral from 0 to 3: \[ V = 2\pi \int_{0}^{3} (5184 - 144x^2 - 48x^4) dx \] Integrate each term: \[ \int_{0}^{3} 5184 dx = 5184x \Big|_0^3 = 15552 \\ \int_{0}^{3} 144x^2 dx = 144 \cdot \frac{x^3}{3} \Big|_0^3 = 144 \cdot 9 = 1296 \\ \int_{0}^{3} 48x^4 dx = 48 \cdot \frac{x^5}{5} \Big|_0^3 = 48 \cdot \frac{243}{5} = \frac{11664}{5} \] Sum: \[ V = 2\pi \left[ 15552 - 1296 - \frac{11664}{5} \right] \] \[ = 2\pi \left[ 14256 - \frac{11664}{5} \right] \] \[ = 2\pi \left[ \frac{71280 - 11664}{5} \right] \] \[ = 2\pi \left[ \frac{59616}{5} \right] \] \[ = \frac{119232\pi}{5} \] **Final Answer:** \[ \boxed{\dfrac{119232\pi}{5}} \] --- ### **6)** **Given:** - \( y = x \), \( y = 3x + 3 \), \( x = 3 \), \( x = 7 \) - Revolve about the y-axis. #### **Step 1: Find bounds for y** For \( x = 3 \), \( y_1 = 3 \), \( y_2 = 12 \) For \( x = 7 \), \( y_1 = 7 \), \( y_2 = 24 \) But, we need to use the shell method because it's about the y-axis. #### **Step 2: Shell method setup** Height of shell: \( (3x + 3) - x = 2x + 3 \) Radius: \( x \) Bounds: from \( x = 3 \) to \( x = 7 \) \[ V = 2\pi \int_{3}^{7} x[(3x + 3) - x] dx = 2\pi \int_{3}^{7} x(2x + 3) dx \] \[ = 2\pi \int_{3}^{7} (2x^2 + 3x) dx \] #### **Step 3: Integrate** \[ \int (2x^2 + 3x) dx = \frac{2}{3}x^3 + \frac{3}{2}x^2 \] Evaluate from 3 to 7: \[ \left[ \frac{2}{3}x^3 + \frac{3}{2}x^2 \right]_3^7 \] At \( x = 7 \): \[ \frac{2}{3} \cdot 343 + \frac{3}{2} \cdot 49 = \frac{686}{3} + \frac{147}{2} \] At \( x = 3 \): \[ \frac{2}{3} \cdot 27 + \frac{3}{2} \cdot 9 = 18 + \frac{27}{2} \] Subtract: \[ \left( \frac{686}{3} + \frac{147}{2} \right) - \left( 18 + \frac{27}{2} \right) \] \[ = \frac{686}{3} + \frac{147-27}{2} - 18 \] \[ = \frac{686}{3} + 60 - 18 \] \[ = \frac{686}{3} + 42 \] Convert 42 to thirds: \[ = \frac{686 + 126}{3} = \frac{812}{3} \] Multiply by \( 2\pi \): \[ V = 2\pi \cdot \frac{812}{3} = \frac{1624\pi}{3} \] **Final Answer:** \[ \boxed{\dfrac{1624\pi}{3}} \] --- ### **7)** **Given:** - \( y = \sqrt{\ln \frac{x}{12}} \) - \( y = \sqrt{\ln \frac{x^2}{144}} \) - \( y = 1 \) - Revolve about x-axis #### **Step 1: Simplify and find bounds** \( y = \sqrt{\ln \frac{x^2}{144}} = \sqrt{2\ln \frac{x}{12}} \) Set \( y = 1 \): \[ 1 = \sqrt{\ln \frac{x}{12}} \implies \ln \frac{x}{12} = 1 \implies \frac{x}{12} = e \implies x = 12e \] \[ 1 = \sqrt{2\ln \frac{x}{12}} \implies 1 = \sqrt{2}\sqrt{\ln \frac{x}{12}} \implies \sqrt{\ln \frac{x}{12}} = \frac{1}{\sqrt{2}} \implies \ln \frac{x}{12} = \frac{1}{2} \implies \frac{x}{12} = e^{1/2} \implies x = 12e^{1/2} \] So, bounds are \( x = 12e^{1/2} \) to \( x = 12e \) #### **Step 2: Volume by Washer Method** Outer radius: \( y_{outer} = \sqrt{2\ln \frac{x}{12}} \) Inner radius: \( y_{inner} = \sqrt{\ln \frac{x}{12}} \) \[ V = \pi \int_{12e^{1/2}}^{12e} \left[ (2\ln \frac{x}{12}) - (\ln \frac{x}{12}) \right] dx \] \[ = \pi \int_{12e^{1/2}}^{12e} \ln \frac{x}{12} dx \] #### **Step 3: Integrate** Let \( u = \ln \frac{x}{12} \), then \( du = \frac{1}{x} dx \), \( dx = x du \), but let's use integration by parts: Let \( f(x) = \ln \frac{x}{12} \), \( dg = dx \). \[ \int \ln \frac{x}{12} dx = x \ln \frac{x}{12} - \int x \cdot \frac{1}{x} dx = x \ln \frac{x}{12} - \int dx = x \ln \frac{x}{12} - x \] Evaluate from \( x = 12e^{1/2} \) to \( x = 12e \): At upper bound \( x = 12e \): \[ x \ln \frac{x}{12} - x = 12e \ln e - 12e = 12e (1) - 12e = 0 \] At lower bound \( x = 12e^{1/2} \): \[ 12e^{1/2} \ln e^{1/2} - 12e^{1/2} = 12e^{1/2} \left(\frac{1}{2}\right) - 12e^{1/2} = 6e^{1/2} - 12e^{1/2} = -6e^{1/2} \] So the definite integral is: \[ 0 - (-6e^{1/2}) = 6e^{1/2} \] So, the volume is: \[ \boxed{6\pi e^{1/2}} \] --- ## **Final Answers:** **5) \( \boxed{\dfrac{119232\pi}{5}} \)** **6) \( \boxed{\dfrac{1624\pi}{3}} \)** **7) \( \boxed{6\pi e^{1/2}} \)** Let me know if you need any clarifications or further breakdowns!