Determine the area of the region bounded by \( x=-y^{2}+10 \) and \( x=(y-2)^{2} \). Determine the area of the region bounded by \( x=y^{2} \) and \( x=12-2 y \) Determine the area of the region bounded by \( y=2 x^{2}+10, y=4 x+1 \) \( x=-2 \) and \( x=5 \).

To find the areas of the regions bounded by the given curves, we will analyze each case separately. ### 1. Area bounded by \( x = -y^2 + 10 \) and \( x = (y - 2)^2 \) First, we need to find the points of intersection of the curves \( x = -y^2 + 10 \) and \( x = (y - 2)^2 \). Setting the equations equal to each other: \[ -y^2 + 10 = (y - 2)^2 \] Expanding the right side: \[ -y^2 + 10 = y^2 - 4y + 4 \] Rearranging gives: \[ 0 = 2y^2 - 4y - 6 \] Dividing by 2: \[ 0 = y^2 - 2y - 3 \] Factoring: \[ 0 = (y - 3)(y + 1) \] Thus, the solutions are \( y = 3 \) and \( y = -1 \). Now, we find the corresponding \( x \) values: For \( y = 3 \): \[ x = -3^2 + 10 = -9 + 10 = 1 \] For \( y = -1 \): \[ x = -(-1)^2 + 10 = -1 + 10 = 9 \] The points of intersection are \( (1, 3) \) and \( (9, -1) \). Next, we set up the integral to find the area between the curves from \( y = -1 \) to \( y = 3 \): \[ \text{Area} = \int_{-1}^{3} \left[(-y^2 + 10) - (y - 2)^2\right] \, dy \] Calculating the integrand: \[ (-y^2 + 10) - (y^2 - 4y + 4) = -y^2 + 10 - y^2 + 4y - 4 = -2y^2 + 4y + 6 \] Now, we compute the integral: \[ \text{Area} = \int_{-1}^{3} (-2y^2 + 4y + 6) \, dy \] Calculating the integral: \[ = \left[-\frac{2}{3}y^3 + 2y^2 + 6y\right]_{-1}^{3} \] Evaluating at the bounds: At \( y = 3 \): \[ -\frac{2}{3}(3^3) + 2(3^2) + 6(3) = -\frac{2}{3}(27) + 2(9) + 18 = -18 + 18 + 18 = 18 \] At \( y = -1 \): \[ -\frac{2}{3}(-1)^3 + 2(-1)^2 + 6(-1) = -\frac{2}{3}(-1) + 2(1) - 6 = \frac{2}{3} + 2 - 6 = \frac{2}{3} - 4 = \frac{2}{3} - \frac{12}{3} = -\frac{10}{3} \] Thus, the area is: \[ \text{Area} = 18 - \left(-\frac{10}{3}\right) = 18 + \frac{10}{3} = \frac{54}{3} + \frac{10}{3} = \frac{64}{3} \] ### 2. Area bounded by \( x = y^2 \) and \( x = 12 - 2y \) Finding the points of intersection: Setting the equations equal: \[ y^2 = 12 - 2y \] Rearranging gives: \[ y^2 + 2y - 12 = 0 \] Factoring: \[ (y + 4)(y - 3) = 0 \] Thus, \( y = -4 \) and \( y = 3 \). Now, we find the corresponding \( x \) values: For \( y = -4 \): \[ x = (-4)^2 = 16 \] For \( y = 3 \): \[ x = 3^2 = 9 \] The points of intersection are \( (16, -4) \) and \( (9, 3) \). Setting up the integral for the area: \[ \text{Area} = \int_{-4}^{3} \left[(12 - 2y) - y^2\right] \, dy \] Calculating the integrand: \[ (12 - 2y) - y^2 = -y^2 - 2y + 12 \] Now, we compute the integral: \[ \text{Area} = \int_{-4}^{3} (-y^2 - 2y + 12) \, dy \] Calculating the integral: \[ = \left[-\frac{1}{3}y^3 - y^2 + 12y\right]_{-4}^{3} \] Evaluating at the bounds: At \( y = 3 \): \[ -\frac{1}{3}(3^3) - (3^2) + 12(3) = -\frac{27}{3} - 9 + 36 = -9 - 9 + 36 = 18 \] At \( y = -4 \): \[ -\frac{1}{3}(-4)^3 - (-4)^2 + 12(-4) = -\frac{1}{3}(-64) - 16 - 48 = \frac{64}{3} - 16 - 48 = \frac{64}{3} - \frac{48}{3} - \frac{144}{3} = \frac{64 - 48 - 144}{3} = \frac{-128}{3} \] Thus, the area is: \[ \text{Area} = 18 - \left(-\frac{128}{3}\right) = 18 + \frac{128}{3} = \frac{54}{3} + \frac{128}{3} = \frac{182}{3} \] ### 3. Area bounded by \( y = 2x^2 + 10 \), \( y = 4x + 1 \), \( x = -2 \), and \( x = 5 \) First, we find the points of intersection of \( y = 2x^2 + 10 \) and \( y = 4x + 1 \): Setting the equations equal: \[ 2x^2 + 10 = 4x + 1 \] Rearranging gives: \[ 2x^2 - 4x + 9 = 0 \] Calculating the discriminant: \[ D = (-4)^2 - 4(2)(9) = 16 - 72 = -56 \] Since the discriminant is negative, there are no real intersections. Thus, we will integrate between the vertical lines \( x = -2 \) and \( x = 5 \). The area is given by: \[ \text{Area} = \int_{-2}^{5} \left[(4x + 1) - (2x^2 + 10)\right] \, dx \] Calculating the integrand: \[ (4x + 1) - (2x^2 + 10) = -2x^2 + 4x - 9 \] Now, we compute the integral: \[ \text{Area} = \int_{-2}^{5} (-2x^2 + 4x - 9) \, dx \] Calculating the integral: \[ = \left[-\frac{2}{3}x^3 + 2x^2 - 9x\right]_{-2}^{5} \] Evaluating at the bounds: At \( x = 5 \): \[ -\frac{2}{3}(5^3) + 2(5^2) - 9(5) = -\frac{2}{3}(125) + 2(25) - 45 = -\frac{250}{3} + 50 - 45 = -\frac{250}{3} + \frac{150}{3} = -\frac{100}{3} \] At \( x = -2 \): \[ -\frac{2}{3}(-2)^3 + 2(-2)^2 - 9(-2) = -\frac{2}{3}(-8) + 2(4) + 18 = \frac{16}{3} + 8 + 18 = \frac{16}{3} + \frac{24}{3} + \frac{54}{3} = \frac{94}{3} \] Thus, the area is: \[ \text{Area} = -\frac{100}{3} - \frac{94}{3} = -\frac{194}{3} \] Since area cannot be negative, we take the absolute value: \[ \text{Area} = \frac{