Calculus 3, Double Integrals

[cosmonautics]

Set up, but do not evaluate, an iterated integral for the volume of the solid.
Under the graph of f(x,y) = 25 - x² - y² and above the xy-plane.

I think f(x,y) looks like this.
I got this double integral. There is a similar problem in my book, but it uses the plane z=16, which made the numbers kind of funny.
Could someone confirm that this (kind of) the correct answer?

 

[Crimson Sunbird]

The \(\displaystyle z\) value goes from 0 to 25. For a value \(\displaystyle z_0\) in between, the intersection of the plane \(\displaystyle z=z_0\) with the surface is the circle with radius \(\displaystyle \sqrt{25-z_0}\); its area is \(\displaystyle \pi(25-z_0)=2\int_{-\sqrt{25-z_0}}^{\sqrt{25-z_0}}\sqrt{25-z_0-x}\,\mathrm dx\). Hence the volume of the solid is \(\displaystyle 2\int_0^{25}\int_{-\sqrt{25-z}}^{\sqrt{25-z}}\sqrt{25-z-x}\,\mathrm dx\,\mathrm dz\).

 

[HallsofIvy]

Set up, but do not evaluate, an iterated integral for the volume of the solid.
Under the graph of f(x,y) = 25 - x² - y² and above the xy-plane.

View attachment 2634I think f(x,y) looks like this.
I got this double integral. There is a similar problem in my book, but it uses the plane z=16, which made the numbers kind of funny.
Could someone confirm that this (kind of) the correct answer?

View attachment 2635

Yes, this is correct. Projecting the paraboloid, \(\displaystyle z= 25- x^2- y^2\), down to the xy-plane gives its base, with z= 0, \(\displaystyle 25- x^2+ y^2\) or \(\displaystyle y^2= 25- x^2\) so that x has a lowest value of -5 and highest value of 5, and, for each x, y goes from \(\displaystyle -\sqrt{25- x^2}\) to \(\displaystyle \sqrt{25- x^2}\).

You may not have had "integrals in polar coordinates" but, since \(\displaystyle x^2+ y^2= 25\) is a circle with center at the origin and radius 5, this could also be done as \(\displaystyle \int_{0}^{2\pi}\int_0^5 (25- r^2)r dr d\theta\).

 

[cosmonautics]

Yes, this is correct. Projecting the paraboloid, \(\displaystyle z= 25- x^2- y^2\), down to the xy-plane gives its base, with z= 0, \(\displaystyle 25- x^2+ y^2\) or \(\displaystyle y^2= 25- x^2\) so that x has a lowest value of -5 and highest value of 5, and, for each x, y goes from \(\displaystyle -\sqrt{25- x^2}\) to \(\displaystyle \sqrt{25- x^2}\).

You may not have had "integrals in polar coordinates" but, since \(\displaystyle x^2+ y^2= 25\) is a circle with center at the origin and radius 5, this could also be done as \(\displaystyle \int_{0}^{2\pi}\int_0^5 (25- r^2)r dr d\theta\).

Thanks so much!