How Do You Solve a Double Integral for Area Between z=e^{x^2} and the xy-Plane?

[tandoorichicken]

I was just faced with this problem on a test and I have no idea how to do it Find the area between the xy-plane and [itex] z= e^{x^2} [/itex] as bounded by x=0, x=1, and y=2x.

I have no idea how to do this problem. I set up the integral as

[tex] \int_{0}^{1} \int_{0}^{2x} e^{x^2} \,dy \,dx [/tex]

 

[arildno]

What is the problem here?
Do the y-integration, and then the x-integration.

 

[M98Ranger]

First, it's important to note that the xy-plane is the plane where z=0. So, we are essentially finding the area between the x-y plane and the curve z=e^{x^2}.

To solve this problem, we can use the concept of double integrals, where we integrate over two variables (in this case, x and y) to find the area between the two curves.

To set up the integral, we start by integrating with respect to y first, since that is the inner integral. The limits of integration for y are from 0 to 2x, as given in the problem.

Next, we integrate with respect to x, with limits of integration from 0 to 1.

So, our integral becomes:

\int_{0}^{1} \int_{0}^{2x} e^{x^2} \,dy \,dx

To solve this, we can use the fundamental theorem of calculus, which states that the integral of a function over a specific interval is equal to the difference of the antiderivatives evaluated at the endpoints of the interval.

In this case, the antiderivative of e^{x^2} is \frac{1}{2}e^{x^2}, so our integral becomes:

\int_{0}^{1} \frac{1}{2}e^{x^2} \,dx

Evaluating this at the endpoints, we get:

\frac{1}{2}e^{1^2} - \frac{1}{2}e^{0^2}

= \frac{1}{2}e - \frac{1}{2}

= \frac{1}{2}(e-1)

So, the area between the xy-plane and z=e^{x^2} bounded by x=0, x=1, and y=2x is \frac{1}{2}(e-1).

It's important to note that this is just one approach to solving this problem. There may be other methods or techniques that can also be used. It's always a good idea to practice different approaches and see which one works best for you.

 

[M98Ranger]

First, it's important to understand what the problem is asking for. In this case, we are finding the area between the xy-plane (z=0) and the function z=e^{x^2}, bounded by the lines x=0, x=1, and y=2x. This means we are looking for the region in the xy-plane where z=0 and z=e^{x^2} intersect, and the boundaries of this region are given by the lines x=0, x=1, and y=2x.

To solve this double integral, we can follow these steps:

1. Draw the region in the xy-plane: Before setting up the integral, it's helpful to visualize the region we are looking for. In this case, it is a triangular region bounded by the lines x=0, x=1, and y=2x. Draw this region on a graph to get a better understanding of the problem.

2. Set up the integral: As you have correctly done, the double integral is set up as \int_{0}^{1} \int_{0}^{2x} e^{x^2} \,dy \,dx. This represents the area under the surface z=e^{x^2} within the given boundaries.

3. Evaluate the inner integral: The inner integral, \int_{0}^{2x} e^{x^2} \,dy, represents the area under the curve e^{x^2} within the boundaries of y=0 and y=2x. This can be solved by simply integrating e^{x^2} with respect to y, giving us e^{x^2}y evaluated from y=0 to y=2x. This simplifies to 2xe^{x^2}.

4. Evaluate the outer integral: The outer integral, \int_{0}^{1} 2xe^{x^2} \,dx, represents the total area of the region bounded by x=0, x=1, and y=2x. This can be solved by integrating 2xe^{x^2} with respect to x, giving us e^{x^2} evaluated from x=0 to x=1. This simplifies to e-1.

5. Final answer: The final answer is e-1, which represents the area between the xy-plane and z=e^{x^2} as bounded by