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I feel like problem number 2 on the exam review should be easy, and I'm just making it hard for no reason. The problem states "Find the volume in the first octant (x, y, z all positive) and under the graph of the function $f(x,y)=4-(x^2 + y^2).$"

With it being in the first octant, $\delta \theta$ should go from 0 to $2\pi$ since we need 1/4 of a circle. And r would go from 0 to 4? This part i'm not exactly sure of, I know that (x^2 + y^2) is a circle, I'm not exactly sure what the 4- goes to, and graphing it on wolfram alpha didn't help much. It looks like it extends it up to 4 in the z direction? So this may be where my original mistake is. So i'm going to work this two ways. The first with $0\leq r \geq 4$ and the second being $ 0\leq r \geq 1$ With my reasoning behind one, because of the $x^2 + y^2$ denoting the unit circle.

$$ \int _0 ^{\frac \pi 2} \int _0 ^4 4-r^2 r \delta r \delta \theta $$ $$ = \frac \pi 2 \int _0 ^4 4-r^3 \delta r$$ $$ = \frac \pi 2 * 4r- \frac 14 r^4 \bigg |_0 ^4 $$ $$ = \frac \pi 2 * 16-64$$ $$ = \frac \pi 2 * -48 $$ $$= \frac {-48\pi}2$$ $$ = -24\pi$$

and the second being

$$ \int _0 ^{\frac \pi 2} \int _0 ^1 4-r^2 r \delta r \delta \theta $$ $$ = \frac \pi 2 \int _0 ^1 4-r^3 \delta r$$ $$ = \frac \pi 2 * 4r- \frac 14 r^4 \bigg |_0 ^1 $$ $$ = \frac \pi 2 * 4-\frac 14$$ $$ = \frac \pi 2 * \frac{15}4 $$ $$= \frac {15\pi}8$$

I'm not sure if either of these are right. So if not, if someone could explain the easiest way to think about this?

I feel like problem number 2 on the exam review should be easy, and I'm just making it hard for no reason. The problem states "Find the volume in the first octant (x, y, z all positive) and under the graph of the function $f(x,y)=4-(x^2 + y^2).$"

With it being in the first octant, $\delta \theta$ should go from 0 to $2\pi$ since we need 1/4 of a circle. And r would go from 0 to 4? This part i'm not exactly sure of, I know that (x^2 $(x^2 + y^2) y^2)$ is a circle, I'm not exactly sure what the 4- goes to, and graphing it on wolfram alpha didn't help much. It looks like it extends it up to 4 in the z direction? So this may be where my original mistake is. So i'm going to work this two ways. The first with $0\leq r \geq 4$ and the second being $ 0\leq r \geq 1$ With my reasoning behind one, because of the $x^2 + y^2$ denoting the unit circle.

$$ \int _0 ^{\frac \pi 2} \int _0 ^4 4-r^2 r \delta r \delta \theta $$ $$ = \frac \pi 2 \int _0 ^4 4-r^3 \delta r$$ $$ = \frac \pi 2 * 4r- \frac 14 r^4 \bigg |_0 ^4 $$ $$ = \frac \pi 2 * 16-64$$ $$ = \frac \pi 2 * -48 $$ $$= \frac {-48\pi}2$$ $$ = -24\pi$$

and the second being

$$ \int _0 ^{\frac \pi 2} \int _0 ^1 4-r^2 r \delta r \delta \theta $$ $$ = \frac \pi 2 \int _0 ^1 4-r^3 \delta r$$ $$ = \frac \pi 2 * 4r- \frac 14 r^4 \bigg |_0 ^1 $$ $$ = \frac \pi 2 * 4-\frac 14$$ $$ = \frac \pi 2 * \frac{15}4 $$ $$= \frac {15\pi}8$$

I'm not sure if either of these are right. So if not, if someone could explain the easiest way to think about this?

I feel like problem number 2 on the exam review should be easy, and I'm just making it hard for no reason. The problem states "Find the volume in the first octant (x, y, z all positive) and under the graph of the function $f(x,y)=4-(x^2 + y^2).$"

With it being in the first octant, $\delta \theta$ should go from 0 to $2\pi$ since we need 1/4 of a circle. And r would go from 0 to 4? This part i'm not exactly sure of, I know that $(x^2 + y^2)$ is a circle, I'm not exactly sure what the 4- goes to, and graphing it on wolfram alpha didn't help much. It looks like it extends it up to 4 in the z direction? So this may be where my original mistake is. So i'm going to work this two ways. The first with $0\leq r \geq \leq 4$ and the second being $ 0\leq r \geq \leq 1$ With my reasoning behind one, because of the $x^2 + y^2$ denoting the unit circle.

$$ \int _0 ^{\frac \pi 2} \int _0 ^4 4-r^2 r \delta r \delta \theta $$ $$ = \frac \pi 2 \int _0 ^4 4-r^3 \delta r$$ $$ = \frac \pi 2 * 4r- \frac 14 r^4 \bigg |_0 ^4 $$ $$ = \frac \pi 2 * 16-64$$ $$ = \frac \pi 2 * -48 $$ $$= \frac {-48\pi}2$$ $$ = -24\pi$$

and the second being

$$ \int _0 ^{\frac \pi 2} \int _0 ^1 4-r^2 r \delta r \delta \theta $$ $$ = \frac \pi 2 \int _0 ^1 4-r^3 \delta r$$ $$ = \frac \pi 2 * 4r- \frac 14 r^4 \bigg |_0 ^1 $$ $$ = \frac \pi 2 * 4-\frac 14$$ $$ = \frac \pi 2 * \frac{15}4 $$ $$= \frac {15\pi}8$$

I'm not sure if either of these are right. So if not, if someone could explain the easiest way to think about this?