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!