이중적분과 삼중적분에서 어려움을 겪는 것은 부피를 구하는 영역 자체를 이해하는 것, 그 이후의 적분구간의 설정과 적분 순서의 결정입니다. 이러한 부분에 ChatGPT가 도움을 줄 수 있습니다. 영역을 주고 ChatGPT에게 답변을 받아봅시다.
[!user] User Enclosed by the cylinders $z=x^2, y=x^2$ and the planes $z=0, y=4$
[!gpt] ChatGPT To visualize the region enclosed by the given surfaces, let’s break them down one by one:
- $z = x^2$: This is a parabolic cylinder opening upwards along the z-axis. If you were to slice the surface perpendicular to the xz-plane, you’d see a parabola.
- $y = x^2$: This is another parabolic cylinder, but this one opens sideways along the y-axis. If you were to slice the surface perpendicular to the xy-plane, you’d see a parabola.
- $z = 0$: This is the xy-plane.
$y = 4$: This is a vertical plane parallel to the xz-plane and located at y = 4. Now, let’s determine the region enclosed by these surfaces:
The region is bounded below by the xy-plane ($z = 0$).
The region is bounded on the sides by the parabolic cylinders $z = x^2$ and $y = x^2$. This means that the region is contained within the “quadrant” formed by these two parabolas.
- The region is bounded on the other side by the plane $y = 4$. This means that the region doesn’t extend indefinitely in the positive y-direction but stops at $y = 4$.
To find the x-values where the region is bounded, set $y = x^2$ equal to 4:
\(x^2 = 4\) \(x = \pm 2\)
So, the region is bounded by $x = 2$ and $x = -2$.
In summary, the enclosed region is:
- Bounded below by the xy-plane.
- Bounded on the sides by the parabolas $z = x^2$ and $y = x^2$.
- Bounded on the top by the portions of the parabolas that lie below the plane $y = 4$.
- Bounded in the x-direction between $x = -2$ and $x = 2$.
If you were to sketch this, you’d see a sort of “curved box” shape, where the sides are formed by the parabolic cylinders and the top is a curved surface formed by the portions of the parabolas that lie below $y = 4$.
ChatGPT가 주어진 영역에 대해 자세히 설명해주었습니다. 하지만 부피를 계산해주지는 않았기 때문에, 부피를 구해달라고 요청해봅시다.
[!user] User how about the volume?
[!gpt] ChatGPT To find the volume of the region enclosed by the cylinders $z = x^2$, $y = x^2$, and the planes $z = 0$, $y = 4$, we can set up a triple integral.
Given the bounds we found earlier:
- $x$ ranges from $-2$ to $2$.
- For a fixed $x$, $y$ ranges from $x^2$ to $4$.
- For a fixed $x$ and $y$, $z$ ranges from $0$ to $x^2$.
The volume $V$ is given by:
\[V = \int_{-2}^{2} \int_{x^2}^{4} \int_{0}^{x^2} dz \, dy \, dx\]The innermost integral with respect to $z$ is straightforward:
\[\int_{0}^{x^2} dz = z \Big|_0^{x^2} = x^2\]Now, integrate with respect to $y$:
\[\int_{x^2}^{4} x^2 \, dy = x^2y \Big|_{x^2}^{4} = 4x^2 - x^4\]Finally, integrate with respect to $x$:
\[V = \int_{-2}^{2} (4x^2 - x^4) \, dx\] \[V = \int_{-2}^{2} 4x^2 \, dx - \int_{-2}^{2} x^4 \, dx\] \[V = \left[ \frac{4}{3}x^3 \right]_{-2}^{2} - \left[ \frac{1}{5}x^5 \right]_{-2}^{2}\] \[V = \left[ \frac{4}{3}(2^3) - \frac{1}{5}(2^5) \right] - \left[ \frac{4}{3}(-2^3) - \frac{1}{5}(-2^5) \right]\] \[V = \left[ \frac{32}{3} - \frac{32}{5} \right] + \left[ \frac{32}{3} + \frac{32}{5} \right]\] \[V = 2\left( \frac{32}{3} - \frac{32}{5} \right)\]To combine the fractions:
\[\frac{32}{3} - \frac{32}{5} = \frac{160}{15} - \frac{96}{15} = \frac{64}{15}\] \[V = 2 \times \frac{64}{15} = \frac{128}{15}\] \[V = \frac{8 \times 16}{15} = \frac{128}{15} = 8\frac{8}{15}\]So, the volume of the region is $8\frac{8}{15}$ cubic units.