A function with two variables can be pictured by making the value of the function a third axis. If the function consists of x and y, the value is z. This turns an area over a plane into a 3 dimensional object. Two integrals are needed to integrate two variable functions defined over a region. The first integral is with respect to one of the variables and the second is with respect to the other. The bounds are the area that is being integrated. Sometimes, that requires the bounds of the inner integral to be a function of the outer variable. Finding the antiderivitive of the function is the same as any other function, only you treat the variable not being integrated as a constant. Once you find the antiderivitive, replace the variable that was integrated with the bounds like you normally would and integrate that new function with respect to the other variable. What you are left with is the value of the function integrated over the area.
The function I chose to integrate is \(\sqrt{y^2-x^2}\). It is integrated over the area bounded by \(y=|x|\) and \(y=4\). This function makes half a cone in the 3d space. In fact, it is the same shape made by rotating \(y=x\) halfway around the y axis. So, the area of the solid could be found by using known cross section, but we're going to do it the other way to learn how to use two integrals. I chose it to show how the large change in slope of a circle. The height of each rectangular prism in each row gets shorter by a larger factor the closer to the outside it is. The height suddenly drops to zero in each row, showing how big of a change the slope is as you get to the edge of the circle.
I approximated the volume by using 20 squares on the xy plane and using the function's value in the top middle of the square as its height. The area of each square is 1, so its value is its volume. The y value of each point goes from 1 to 4 and the x value goes from \(-(y-.5)\)to \(y-.5\) both with increments of 1. A formula to find the volume of the approximation is \[\sum_{y=1}^4{\sum_{x=-y+1}^{y}{\sqrt{y^2-(x-.5)^2}}}\] This comes to equal 48.151.
Because the lower bounds of my area is |x|, my integration needs to be split into two parts. One with the inner integral bounds as x and 4 and one with the bounds as -x and 4. The function and the bounds are symmetric, though, so each of the parts equals one half of the actual value. So, we can just do one of the parts and multiply it by 2. The two integrals required are \[2*\int_{0}^4{\int_x^4{\sqrt{y^2-x^2}}dy}dx\] Finding the antiderivitive of the function and replacing the y's with the the bounds gets you \[2*\int_0^4{(.5(4\sqrt{4^2-x^2}-x^2\ln{(\sqrt{4^2-x^2}+4)})-.5(-x^2\ln{x}))}dx\] Finally, solving that integral and multiplying by 2 gets you 33.51. It makes sense that the approximation's value is bigger than the actual value because the rectangular prisms in the approximation had heights that were taller than the average height each square would have in the actual solid.
The function I chose to integrate is \(\sqrt{y^2-x^2}\). It is integrated over the area bounded by \(y=|x|\) and \(y=4\). This function makes half a cone in the 3d space. In fact, it is the same shape made by rotating \(y=x\) halfway around the y axis. So, the area of the solid could be found by using known cross section, but we're going to do it the other way to learn how to use two integrals. I chose it to show how the large change in slope of a circle. The height of each rectangular prism in each row gets shorter by a larger factor the closer to the outside it is. The height suddenly drops to zero in each row, showing how big of a change the slope is as you get to the edge of the circle.
I approximated the volume by using 20 squares on the xy plane and using the function's value in the top middle of the square as its height. The area of each square is 1, so its value is its volume. The y value of each point goes from 1 to 4 and the x value goes from \(-(y-.5)\)to \(y-.5\) both with increments of 1. A formula to find the volume of the approximation is \[\sum_{y=1}^4{\sum_{x=-y+1}^{y}{\sqrt{y^2-(x-.5)^2}}}\] This comes to equal 48.151.
Because the lower bounds of my area is |x|, my integration needs to be split into two parts. One with the inner integral bounds as x and 4 and one with the bounds as -x and 4. The function and the bounds are symmetric, though, so each of the parts equals one half of the actual value. So, we can just do one of the parts and multiply it by 2. The two integrals required are \[2*\int_{0}^4{\int_x^4{\sqrt{y^2-x^2}}dy}dx\] Finding the antiderivitive of the function and replacing the y's with the the bounds gets you \[2*\int_0^4{(.5(4\sqrt{4^2-x^2}-x^2\ln{(\sqrt{4^2-x^2}+4)})-.5(-x^2\ln{x}))}dx\] Finally, solving that integral and multiplying by 2 gets you 33.51. It makes sense that the approximation's value is bigger than the actual value because the rectangular prisms in the approximation had heights that were taller than the average height each square would have in the actual solid.
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