Let's break down the problem step by step: ### **Step 1: Understand the Region** The dam wall is the region above \( y = 0.8x^2 \) and below \( y = 240 \). - The **top** of the dam: \( y = 240 \) - The **bottom** of the dam: where \( y = 0.8x^2 \) - The **width** at any \( y \) is determined by solving \( y = 0.8x^2 \) for \( x \): \[ y = 0.8x^2 \implies x^2 = \frac{y}{0.8} \implies x = \pm \sqrt{\frac{y}{0.8}} \] So, at height \( y \), the width is: \[ \text{Width}(y) = 2\sqrt{\frac{y}{0.8}} \] ### **Step 2: Water Level** - Water is 32 meters below the top: \( y = 240 - 32 = 208 \) - Water exerts force on the region \( 208 \leq y \leq 240 \) ### **Step 3: Pressure at Depth** At depth \( h \) below the water surface, pressure is \( P = \rho g h \). - At height \( y \), depth below water surface: \( h = 240 - y \) - So, pressure at height \( y \): \( P = 1000 \cdot 9.8 \cdot (240 - y) \) ### **Step 4: Strip Area** A thin horizontal strip at height \( y \), thickness \( dy \): - Area \( dA = \text{Width}(y) \cdot dy = 2\sqrt{\frac{y}{0.8}} \, dy \) ### **Step 5: Force on Strip** Force on strip: \( dF = P \cdot dA = 1000 \cdot 9.8 \cdot (240 - y) \cdot 2\sqrt{\frac{y}{0.8}} \, dy \) \[ dF = 1000 \cdot 9.8 \cdot 2 \cdot (240 - y) \sqrt{\frac{y}{0.8}} \, dy \] ### **Step 6: Total Force** Integrate from \( y = 208 \) to \( y = 240 \): \[ F = \int_{208}^{240} 1000 \cdot 9.8 \cdot 2 \cdot (240 - y) \sqrt{\frac{y}{0.8}} \, dy \] \[ F = 19,600 \int_{208}^{240} (240 - y) \sqrt{\frac{y}{0.8}} \, dy \] \[ \sqrt{\frac{y}{0.8}} = \frac{1}{\sqrt{0.8}} y^{1/2} \] So, \[ F = 19,600 \cdot \frac{1}{\sqrt{0.8}} \int_{208}^{240} (240 - y) y^{1/2} \, dy \] Calculate \( \frac{1}{\sqrt{0.8}} \approx 1.118 \): \[ F \approx 21,912.8 \int_{208}^{240} (240 - y) y^{1/2} \, dy \] ### **Step 7: Compute the Integral** Let’s expand: \[ \int_{208}^{240} (240 - y) y^{1/2} \, dy = 240 \int_{208}^{240} y^{1/2} \, dy - \int_{208}^{240} y^{3/2} \, dy \] Compute each: 1. \( \int y^{1/2} dy = \frac{2}{3} y^{3/2} \) 2. \( \int y^{3/2} dy = \frac{2}{5} y^{5/2} \) Plug in: \[ 240 \cdot \frac{2}{3} y^{3/2} \Bigg|_{208}^{240} - \frac{2}{5} y^{5/2} \Bigg|_{208}^{240} \] Calculate values: - \( 240^{1/2} = 15.4919 \), \( 240^{3/2} = 240 \cdot 15.4919 = 3,718.06 \) - \( 208^{1/2} = 14.4222 \), \( 208^{3/2} = 208 \cdot 14.4222 = 3,001.82 \) - \( 240^{5/2} = 240^{2} \cdot 240^{1/2} = 57,600 \cdot 15.4919 = 892,293.44 \) - \( 208^{5/2} = 208^2 \cdot 14.4222 = 43,264 \cdot 14.4222 = 623,772.38 \) Plug in: \[ = 240 \cdot \frac{2}{3} (3,718.06 - 3,001.82) - \frac{2}{5}(892,293.44 - 623,772.38) \] \[ = 240 \cdot \frac{2}{3} \cdot 716.24 - \frac{2}{5} \cdot 268,521.06 \] \[ = 240 \cdot 477.493 - 107,408.424 \] \[ = 114,598.32 - 107,408.424 = 7,189.896 \] ### **Step 8: Final Force** \[ F \approx 21,912.8 \times 7,189.896 \approx 157,495,654 \text{ Newtons} \] --- ## **Final Answer** \[ \boxed{1.57 \times 10^8 \text{ N}} \] The force exerted on the dam by the water is approximately **157,000,000 Newtons**.