I would integrate with respect to y, becaus if you did it with respect to x
x=+/- sqrot(30-y^2) is not a function but a relation.
Am I right?
1. Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
y= sqrt(30-x) and y=-x i inserted in graphing calculator but no graph???
2. Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
I graphed and is from 0 to 3.
y= 4x^3/3 and y=1/3x(x^2+9)
i got 54. did i do it right?
You don't need a graphing calculator for this. It's best to have your equations for the first problem as . So we have and Set the two equal to each other to find the points of intersection:
You should be able to sketch of graph of this and know that on the interval of , so your integral should be:
For the second problem, it's obvious we should have our functions as . Again, set the two equations equal to each other:
Again, sketching the graphs of these are easy, and you can tell on , so your integral would be: