Finding volumes via double integrals

In summary, the question asks to find the volume below the plane z = 2x + 3y, bounded by the x- and y-axes and the line x + y = 1 in the x-y plane. This can be solved by slicing the volume with vertical slices parallel to the y-axis and integrating the area of each slice. The limits of integration are 0 <= x <= 1 and -x <= y <= x.
  • #1
lmstaples
31
0

Homework Statement



Find the volume which lies below the plane [itex]z = 2x + 3y[/itex] and whose base in the [itex]x - y[/itex] plane is bounded by the [itex]x-[/itex] and [itex]y-[/itex]axes and the line [itex]x + y = 1[/itex].


Homework Equations



[itex]I = \int\int_{R} f(x, y) dydx = \int^{b}_{a}\int^{y=y_{2}(x)}_{y=y_{1}(x)} f(x, y) dydx[/itex]


The Attempt at a Solution



I know the integral is of the form:

[itex]\int\int_{C} (2x + 3y) dydx[/itex]

But I am very confused which what the limits are as they give so much information!

I thought maybe something along the lines:

[itex]x = 1 - y \Rightarrow 0 = 2(1 - y) + 3y = 2 - 2y + 3y = 2 + y \Rightarrow y = -2 \Rightarrow x = 3[/itex] (not sure if this helps in any way at all)

And : [itex]2x + 3y = 0 \Rightarrow 3y = -2x \Rightarrow y = -\frac{2}{3}x[/itex] (but once again not sure if this even means anything)

I haven't been formally taught this (at least finding volumes I haven't - I have had 2 tutorials on simply integrating over a region R) and expected to simply DO the assignment :(

Thanks for any help!
 
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  • #2
Consider slicing the volume with vertical slices parallel to the [itex]y[/itex]-axis, which results in this integration:
[tex]V = \int A(x) dx[/tex]
where [itex]A(x)[/itex] is the area of each slice. To do the integration, first find an expression for [itex]A(x)[/itex] in term of [itex]x[/itex].
 
  • #3
I'm really sorry but I have no idea what you are asking for :(

Would the limits be 0 <= x <= 1

-x <= y <= x

?
 
  • #4
Look up "volume by slicing".
 

FAQ: Finding volumes via double integrals

What is a double integral?

A double integral is a mathematical tool used to calculate the volume under a surface in three-dimensional space.

What is the difference between a single integral and a double integral?

A single integral calculates the area under a curve in two-dimensional space, while a double integral calculates the volume under a surface in three-dimensional space.

How do you set up a double integral to find volume?

To set up a double integral for finding volume, you first need to determine the limits of integration for both the x and y variables. This will create a rectangular region on the x-y plane. Then, you need to determine the function that represents the surface and set it up as the integrand. Finally, you integrate the function over the region to find the volume.

What is the importance of finding volumes via double integrals?

Finding volumes via double integrals is an important tool in mathematics and science, as it allows us to calculate the volume of complex three-dimensional objects and surfaces. This can be applied in various fields such as physics, engineering, and economics.

What are some real-life applications of finding volumes via double integrals?

Some real-life applications of finding volumes via double integrals include calculating the volume of a liquid in a container, determining the mass of an irregularly shaped object, and calculating the surface area of a three-dimensional object for manufacturing purposes.

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