Triple Integrals

Question

I have below problem on which I'm working on now.

Could someone explain how do I solve this problem?

Question and my work: https://drive(dot)google.com/file/d/1-3EOv0RBOfDikMFGFwzRGs0zYBiVRNmi/view?usp=drivesdk

Answer 1

Use this website and copy and paste the solution:.....

quicklatex.com

**Problem.** Compute the volume of the solid bounded above by

\[

z = 4 - x

\]

and below by

\[

z = 2x^2 + y^2.

\]

---

### 1. Find the boundary in the \(xy\)–plane

Set

\[

4 - x = 2x^2 + y^2

\;\Longrightarrow\;

2x^2 + x + y^2 - 4 = 0.

\]

Complete the square in \(x\):

\[

2\!\bigl(x^2 + \tfrac12x\bigr) + y^2 = 4

\quad\Rightarrow\quad

2\!\Bigl[(x + \tfrac14)^2 - \tfrac1{16}\Bigr] + y^2 = 4

\]

\[

\Longrightarrow\;

2(x + \tfrac14)^2 + y^2 = 4 + \tfrac18 = \tfrac{33}{8}.

\]

Divide through by \(\tfrac{33}{8}\) to get the standard ellipse:

\[

\frac{(x + \tfrac14)^2}{\tfrac{33}{16}} + \frac{y^2}{\tfrac{33}{8}} = 1.

\]

So the region \(D\) is an ellipse centered at \(\bigl(-\tfrac14,0\bigr)\) with semi-axes

\[

a = \sqrt{\tfrac{33}{16}} = \frac{\sqrt{33}}{4},

\quad

b = \sqrt{\tfrac{33}{8}} = \frac{\sqrt{33}}{2\sqrt2}.

\]

---

### 2. Set up the volume integral

\[

V \;=\;\iint_D \bigl[(4 - x) - (2x^2 + y^2)\bigr]\,dA

\;=\;\iint_D \bigl[4 - x - 2x^2 - y^2\bigr]\,dx\,dy.

\]

**Shift coordinates**: let

\[

u = x + \tfrac14,\quad v = y,

\]

so that \(D\) becomes \(\tfrac{u^2}{a^2} + \tfrac{v^2}{b^2}\le1\), and

\[

x = u - \tfrac14,

\quad

4 - x - 2x^2 - y^2

= 4 - \bigl(u-\tfrac14\bigr) - 2\bigl(u-\tfrac14\bigr)^2 - v^2

= \frac{33}{8} - 2u^2 - v^2.

\]

The Jacobian is \(dA = du\,dv\).

---

### 3. Switch to “stretched” polar in \((u,v)\)

Use

\[

u = a\,r\cos\theta,\quad

v = b\,r\sin\theta,

\quad

0 \le r \le 1,\; 0\le\theta\le2\pi,

\]

with \(a=\sqrt{33}/4,\; b=\sqrt{33}/(2\sqrt2)\). Then \(dA = (ab)\,r\,dr\,d\theta\).

The integrand simplifies:

\[

\frac{33}{8} - 2u^2 - v^2

= \frac{33}{8} - 2(a^2r^2\cos^2\theta) - (b^2r^2\sin^2\theta)

= \frac{33}{8}\bigl(1 - r^2\bigr),

\]

because \(2a^2 = \tfrac{33}{8}\) and \(b^2 = \tfrac{33}{8}\).

Hence

\[

V

= \int_{0}^{2\pi}\!\int_{0}^{1}

\frac{33}{8}(1 - r^2)\,(ab)\,r\,dr\,d\theta

= ab\;\frac{33}{8}\;2\pi\;\int_{0}^{1}(r - r^3)\,dr.

\]

Compute the radial integral:

\[

\int_{0}^{1}(r - r^3)\,dr = \Bigl[\tfrac12r^2 - \tfrac14r^4\Bigr]_0^1

= \tfrac12 - \tfrac14 = \tfrac14.

\]

So

\[

V = ab\;\frac{33}{8}\;2\pi\;\frac14

= ab\;\frac{33\pi}{16}.

\]

Finally, \(ab = \tfrac{\sqrt{33}}{4}\cdot\tfrac{\sqrt{33}}{2\sqrt2} = \tfrac{33}{8\sqrt2}\), giving

\[

V = \frac{33}{8\sqrt2}\,\frac{33\pi}{16}

= \frac{1089\pi}{128\sqrt2}

= \frac{1089\sqrt2\,\pi}{256}.

\]

---

**Answer.**

\[

\boxed{

V = \frac{1089\sqrt2\,\pi}{256}

}

\]