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### 5b. Evaluate the Double Integral [tex]$\int_0^1 \int_0^2 (x^2 + y^2) \, dA$[/tex]
To evaluate the given double integral, follow these steps:
1. Set Up the Double Integral:
[tex]\[ \int_{0}^{1} \int_{0}^{2} (x^2 + y^2) \, dx \, dy \][/tex]
2. Integrate with Respect to x First:
Consider the inner integral:
[tex]\[ \int_{0}^{2} (x^2 + y^2) \, dx \][/tex]
This can be broken down as:
[tex]\[ \int_{0}^{2} x^2 \, dx + \int_{0}^{2} y^2 \, dx \][/tex]
3. Integrate [tex]$x^2$[/tex] with Respect to [tex]$x$[/tex]:
[tex]\[ \int_{0}^{2} x^2 \, dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \][/tex]
4. Integrate [tex]$y^2$[/tex] with Respect to [tex]$x$[/tex]:
Since [tex]$y^2$[/tex] is a constant with respect to [tex]$x$[/tex], it comes out of the integral:
[tex]\[ \int_{0}^{2} y^2 \, dx = y^2 \int_{0}^{2} 1 \, dx = y^2 [x]_0^2 = y^2 (2 - 0) = 2y^2 \][/tex]
5. Combine the Results:
[tex]\[ \int_{0}^{2} (x^2 + y^2) \, dx = \frac{8}{3} + 2y^2 \][/tex]
6. Integrate with Respect to [tex]$y$[/tex] Next:
Now consider the outer integral:
[tex]\[ \int_{0}^{1} \left(\frac{8}{3} + 2y^2\right) \, dy \][/tex]
7. Integrate [tex]$\frac{8}{3}$[/tex] with Respect to [tex]$y$[/tex]:
[tex]\[ \int_{0}^{1} \frac{8}{3} \, dy = \frac{8}{3} \left[y\right]_0^1 = \frac{8}{3} (1 - 0) = \frac{8}{3} \][/tex]
8. Integrate [tex]$2y^2$[/tex] with Respect to [tex]$y$[/tex]:
[tex]\[ \int_{0}^{1} 2y^2 \, dy = 2 \int_{0}^{1} y^2 \, dy = 2 \left[\frac{y^3}{3}\right]_0^1 = 2 \cdot \frac{1^3}{3} = \frac{2}{3} \][/tex]
9. Combine the Results:
[tex]\[ \int_{0}^{1} \int_{0}^{2} (x^2 + y^2) \, dA = \frac{8}{3} + \frac{2}{3} = \frac{10}{3} \][/tex]
Thus, the value of the double integral is given by:
[tex]\[ \int_{0}^{1} \int_{0}^{2} (x^2 + y^2) \, dA = \frac{10}{3} \][/tex]
### 6. Find the Angle Between the Surface [tex]$x^2 + y^2 + z^2 = 9$[/tex] and [tex]$z = x^2 + y^2 - 3$[/tex] at [tex]$(2, -1, 2)$[/tex]
To find the angle between two surfaces at a given point, we need to compute the angle between their normal vectors at that point. The normal vector to a surface defined by [tex]$f(x, y, z) = 0$[/tex] is given by the gradient [tex]$\nabla f$[/tex].
1. Find the Gradient of the First Surface [tex]$x^2 + y^2 + z^2 = 9$[/tex]:
[tex]\[ \nabla f = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) (x^2 + y^2 + z^2) = (2x, 2y, 2z) \][/tex]
At the point [tex]$(2, -1, 2)$[/tex], the gradient is:
[tex]\[ \nabla f(2, -1, 2) = (2 \cdot 2, 2 \cdot -1, 2 \cdot 2) = (4, -2, 4) \][/tex]
2. Find the Gradient of the Second Surface [tex]$z = x^2 + y^2 - 3$[/tex]:
Rewrite this as [tex]$g(x, y, z) = x^2 + y^2 - z - 3 = 0$[/tex] and find its gradient:
[tex]\[ \nabla g = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) (x^2 + y^2 - z - 3) = (2x, 2y, -1) \][/tex]
At the point [tex]$(2, -1, 2)$[/tex], the gradient is:
[tex]\[ \nabla g(2, -1, 2) = (2 \cdot 2, 2 \cdot -1, -1) = (4, -2, -1) \][/tex]
3. Calculate the Dot Product of the Two Normal Vectors:
[tex]\[ \nabla f \cdot \nabla g = (4, -2, 4) \cdot (4, -2, -1) = 4 \cdot 4 + (-2) \cdot (-2) + 4 \cdot (-1) = 16 + 4 - 4 = 16 \][/tex]
4. Calculate the Magnitudes of the Normal Vectors:
[tex]\[ |\nabla f| = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \][/tex]
[tex]\[ |\nabla g| = \sqrt{4^2 + (-2)^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21} \][/tex]
5. Use the Dot Product to Find the Cosine of the Angle [tex]$\theta$[/tex]:
[tex]\[ \cos \theta = \frac{\nabla f \cdot \nabla g}{|\nabla f| |\nabla g|} = \frac{16}{6 \sqrt{21}} = \frac{16}{6 \sqrt{21}} = \frac{8}{3 \sqrt{21}} \][/tex]
6. Find the Actual Angle:
[tex]\[ \theta = \cos^{-1} \left( \frac{8}{3 \sqrt{21}} \right) \][/tex]
Therefore, the angle between the surfaces [tex]$x^2 + y^2 + z^2 = 9$[/tex] and [tex]$z = x^2 + y^2 - 3$[/tex] at the point [tex]$(2, -1, 2)$[/tex] is:
[tex]\[ \theta = \cos^{-1} \left( \frac{8}{3 \sqrt{21}} \right) \][/tex]
### (ib) Find the Limit of the Following Functions:
You didn't specify the functions for which the limits need to be found, but here's the general approach:
1. Identify the Function and the Point of Interest:
Determine the function [tex]$f(x)$[/tex] or [tex]$f(x, y)$[/tex] for which you need to find the limit as [tex]$x \to a$[/tex] or [tex]$(x, y) \to (a, b)$[/tex].
2. Evaluate the Limit:
- For one-variable functions, directly substitute the point into the function if it is continuous at that point.
- For multivariable functions, consider approaching the point from different paths to ensure the limit is the same from all directions.
If you provide specific functions, I can give you a detailed step-by-step solution for those limits.
### 5b. Evaluate the Double Integral [tex]$\int_0^1 \int_0^2 (x^2 + y^2) \, dA$[/tex]
To evaluate the given double integral, follow these steps:
1. Set Up the Double Integral:
[tex]\[ \int_{0}^{1} \int_{0}^{2} (x^2 + y^2) \, dx \, dy \][/tex]
2. Integrate with Respect to x First:
Consider the inner integral:
[tex]\[ \int_{0}^{2} (x^2 + y^2) \, dx \][/tex]
This can be broken down as:
[tex]\[ \int_{0}^{2} x^2 \, dx + \int_{0}^{2} y^2 \, dx \][/tex]
3. Integrate [tex]$x^2$[/tex] with Respect to [tex]$x$[/tex]:
[tex]\[ \int_{0}^{2} x^2 \, dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \][/tex]
4. Integrate [tex]$y^2$[/tex] with Respect to [tex]$x$[/tex]:
Since [tex]$y^2$[/tex] is a constant with respect to [tex]$x$[/tex], it comes out of the integral:
[tex]\[ \int_{0}^{2} y^2 \, dx = y^2 \int_{0}^{2} 1 \, dx = y^2 [x]_0^2 = y^2 (2 - 0) = 2y^2 \][/tex]
5. Combine the Results:
[tex]\[ \int_{0}^{2} (x^2 + y^2) \, dx = \frac{8}{3} + 2y^2 \][/tex]
6. Integrate with Respect to [tex]$y$[/tex] Next:
Now consider the outer integral:
[tex]\[ \int_{0}^{1} \left(\frac{8}{3} + 2y^2\right) \, dy \][/tex]
7. Integrate [tex]$\frac{8}{3}$[/tex] with Respect to [tex]$y$[/tex]:
[tex]\[ \int_{0}^{1} \frac{8}{3} \, dy = \frac{8}{3} \left[y\right]_0^1 = \frac{8}{3} (1 - 0) = \frac{8}{3} \][/tex]
8. Integrate [tex]$2y^2$[/tex] with Respect to [tex]$y$[/tex]:
[tex]\[ \int_{0}^{1} 2y^2 \, dy = 2 \int_{0}^{1} y^2 \, dy = 2 \left[\frac{y^3}{3}\right]_0^1 = 2 \cdot \frac{1^3}{3} = \frac{2}{3} \][/tex]
9. Combine the Results:
[tex]\[ \int_{0}^{1} \int_{0}^{2} (x^2 + y^2) \, dA = \frac{8}{3} + \frac{2}{3} = \frac{10}{3} \][/tex]
Thus, the value of the double integral is given by:
[tex]\[ \int_{0}^{1} \int_{0}^{2} (x^2 + y^2) \, dA = \frac{10}{3} \][/tex]
### 6. Find the Angle Between the Surface [tex]$x^2 + y^2 + z^2 = 9$[/tex] and [tex]$z = x^2 + y^2 - 3$[/tex] at [tex]$(2, -1, 2)$[/tex]
To find the angle between two surfaces at a given point, we need to compute the angle between their normal vectors at that point. The normal vector to a surface defined by [tex]$f(x, y, z) = 0$[/tex] is given by the gradient [tex]$\nabla f$[/tex].
1. Find the Gradient of the First Surface [tex]$x^2 + y^2 + z^2 = 9$[/tex]:
[tex]\[ \nabla f = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) (x^2 + y^2 + z^2) = (2x, 2y, 2z) \][/tex]
At the point [tex]$(2, -1, 2)$[/tex], the gradient is:
[tex]\[ \nabla f(2, -1, 2) = (2 \cdot 2, 2 \cdot -1, 2 \cdot 2) = (4, -2, 4) \][/tex]
2. Find the Gradient of the Second Surface [tex]$z = x^2 + y^2 - 3$[/tex]:
Rewrite this as [tex]$g(x, y, z) = x^2 + y^2 - z - 3 = 0$[/tex] and find its gradient:
[tex]\[ \nabla g = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) (x^2 + y^2 - z - 3) = (2x, 2y, -1) \][/tex]
At the point [tex]$(2, -1, 2)$[/tex], the gradient is:
[tex]\[ \nabla g(2, -1, 2) = (2 \cdot 2, 2 \cdot -1, -1) = (4, -2, -1) \][/tex]
3. Calculate the Dot Product of the Two Normal Vectors:
[tex]\[ \nabla f \cdot \nabla g = (4, -2, 4) \cdot (4, -2, -1) = 4 \cdot 4 + (-2) \cdot (-2) + 4 \cdot (-1) = 16 + 4 - 4 = 16 \][/tex]
4. Calculate the Magnitudes of the Normal Vectors:
[tex]\[ |\nabla f| = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \][/tex]
[tex]\[ |\nabla g| = \sqrt{4^2 + (-2)^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21} \][/tex]
5. Use the Dot Product to Find the Cosine of the Angle [tex]$\theta$[/tex]:
[tex]\[ \cos \theta = \frac{\nabla f \cdot \nabla g}{|\nabla f| |\nabla g|} = \frac{16}{6 \sqrt{21}} = \frac{16}{6 \sqrt{21}} = \frac{8}{3 \sqrt{21}} \][/tex]
6. Find the Actual Angle:
[tex]\[ \theta = \cos^{-1} \left( \frac{8}{3 \sqrt{21}} \right) \][/tex]
Therefore, the angle between the surfaces [tex]$x^2 + y^2 + z^2 = 9$[/tex] and [tex]$z = x^2 + y^2 - 3$[/tex] at the point [tex]$(2, -1, 2)$[/tex] is:
[tex]\[ \theta = \cos^{-1} \left( \frac{8}{3 \sqrt{21}} \right) \][/tex]
### (ib) Find the Limit of the Following Functions:
You didn't specify the functions for which the limits need to be found, but here's the general approach:
1. Identify the Function and the Point of Interest:
Determine the function [tex]$f(x)$[/tex] or [tex]$f(x, y)$[/tex] for which you need to find the limit as [tex]$x \to a$[/tex] or [tex]$(x, y) \to (a, b)$[/tex].
2. Evaluate the Limit:
- For one-variable functions, directly substitute the point into the function if it is continuous at that point.
- For multivariable functions, consider approaching the point from different paths to ensure the limit is the same from all directions.
If you provide specific functions, I can give you a detailed step-by-step solution for those limits.
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