Calculus III i need help

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mathb0bita

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I have been trying to solve this problem for hours. Can't seem to figure out how. Please help me.

IMG_0966.JPG
 
mathb0bita said:
I have been trying to solve this problem for hours. Can't seem to figure out how. Please help me.

View attachment 27018
Click to expand...
Since you have been trying, you presumably have at least one failed attempt to show; we need that, in order to see what is going wrong (along with an explanation of what is bothering you about it).

But one possibility is that the problem is not entirely clear. You will be integrating over a region in the xy-plane that is bounded by the four curves shown here:

1620520309590.png

But which region? There are many regions here, but one that is bounded by all four appears to be this:

1620520610282.png

(There's another possibility on the left.)

If I'm right, you'll have to split the integral into two parts. That's an ugly problem, so I hope I'm reading something wrong!
 
Dr.Peterson said:
Since you have been trying, you presumably have at least one failed attempt to show; we need that, in order to see what is going wrong (along with an explanation of what is bothering you about it).

But one possibility is that the problem is not entirely clear. You will be integrating over a region in the xy-plane that is bounded by the four curves shown here:

View attachment 27030

But which region? There are many regions here, but one that is bounded by all four appears to be this:

View attachment 27031

(There's another possibility on the left.)

If I'm right, you'll have to split the integral into two parts. That's an ugly problem, so I hope I'm reading something wrong!
Click to expand...
Sorry it's my first time posting on this forum. But yes thank you for the help the paraboloid part was what confused me. Just figured out that it would be the function that should be integrated.
 
Yes, the region is not well-defined. It could equally be the region on the left. I suppose you could easily do both!
 
Assuming that the region is the one Dr. Peterson shaded in post #3, then x must go from 0 to 2 but we need to do that in two different parts. For x from 0 to 1, y goes from the line y= x on the bottom to y= 5 on top. For x from 1 to 2, y goes from y= x on the bottom to \(\displaystyle y= 6- x^2\) on the top. The function to be integrated is \(\displaystyle z= x^2+y^2\) so the volume is given by \(\displaystyle \int_{x=0}^1\int_{y=x}^5 (x^2+ y^2) dydx+ \int_{x= 1}^2\int_{y=x}^{6- x^2} (x^2+ y^2) dydx\)
 
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