QUESTION 4 [TOTAL MARKS: 5 MARKS] A manufacturing company wants to calculate the volume of a cylindrical fuel tank with a parabolic cross-section defined by \( y=x^{2} \) and bounded between \( x=-1 \) and \( x=1 \). Using integration, evaluate the volume of the tank. [5 marks]

To find the volume of the cylindrical fuel tank with a parabolic cross-section defined by \( y=x^{2} \) and bounded between \( x=-1 \) and \( x=1 \), we can use the formula for the volume of a solid of revolution: \[ V = \pi \int_{a}^{b} [f(x)]^2 dx \] where: - \( V \) is the volume of the solid, - \( f(x) \) is the function that defines the cross-section of the solid, - \( a \) and \( b \) are the limits of integration. In this case, the function that defines the cross-section is \( y=x^{2} \), and the limits of integration are \( x=-1 \) and \( x=1 \). Substitute the function and limits into the formula: \[ V = \pi \int_{-1}^{1} (x^{2})^2 dx \] Now, we can evaluate the integral to find the volume of the tank. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\pi \times \int_{-1}^{1} \left(x^{2}\right)^{2} dx\) - step1: Multiply the exponents: \(\pi \times \int_{-1}^{1} x^{2\times 2} dx\) - step2: Multiply the numbers: \(\pi \times \int_{-1}^{1} x^{4} dx\) - step3: Evaluate the integral: \(\pi \times \frac{2}{5}\) - step4: Multiply: \(\frac{\pi \times 2}{5}\) - step5: Multiply: \(\frac{2\pi }{5}\) The volume of the cylindrical fuel tank with a parabolic cross-section defined by \( y=x^{2} \) and bounded between \( x=-1 \) and \( x=1 \) is \( \frac{2\pi}{5} \) cubic units.