Couple of ways to go here:

We know sec = 1/cos

Use calculator to find the angle....then find the cos...then find  1/cos

    Angle is 20.5546 degrees   cos=.93633   1/cos= sec = 1.068

Remember sin^2 + cos^2=1   solve for cos then 1/cos

      1-.3511^2 = cos^2   cos=.93633    then 1/cos = sec = 1.068

Thanks, Alan.

Thank you GingerAle, I like to plot graphs where possible - they often make it easier to see what is going on.

The integral Lancelot link gives after "V = " is the one I've written below my 3D graph. Lancelot provided more explicit steps to the solution than I did (I skipped over a few!).

It is sad that this forum has such a poor search facility.  An alternative way of searching it is to use your browser's search facility.  From outside the forum enter something like: "web2.0calc Badinage xxx" where you replace xxx with a key word or two from the question you are trying to find. Not ideal, but with luck you might have more success than with the forum search engine.

Sir Alan, your graph makes this come alive.

I do not know how to do double integrals (yet), but last year a second-year uni-mate asked me to forward this exact question to Lancelot Link.  This is his solution minus his snarky comments about how his monkey-barkeeper, Stu, could answer it.  (This is really true. I didn’t make that comment to troll Miss Melody or Sir Alan – even if it does seem like it. LOL)  

I made only minor changes to the Latex code so it will properly display on the forum.

\(\text {Using the given points, graph the triangle. Note that x lies from 1 to 4 inclusive, } \\ \text {and y lies from 1 to the line connecting the points (1,2); (4.1). }\\ \text {By inspection, this region is a type I. This gives the equation: }\\ \)\(y-2 = - \dfrac{1}{3}(x-1)\\ y= \dfrac{1}{3}x + \dfrac{7}{3}\\ Set \, D=(x, y)\, | 1\leq x\leq 4,\, 1\leq y\leq {\frac{-x}{3}+\frac{7}{3}}\\ V = \int_{1}^{4}\int_{1}^{\frac{-x}{3}+\frac{7}{3}} xy {\, } dydx = \int_{1}^{4}\, \left [\dfrac {xy^2}{2} \right ]_1^{\frac{-x}{3}+\frac{7}{3}}\, dx = \frac{1}{2} \int_{1}^{4}\, \left ( x \left ( -\dfrac{x}{3}+ \dfrac{7}{3}\right )^{2}-x \right )dx \\ = \dfrac{1}{18}\int_{1}^{4}\, \left (x^3 - 14x^2 +40x \right )dx = \dfrac{1}{18} \left [ \dfrac{x^4}{4}- \dfrac{14x^3}{3}+20x^2 \right ]_1^4 \\ = \dfrac{1}{18} \left (\dfrac{1}{4} (255)-\frac{14}{3}(63)+20(15) \right )\\ = \dfrac{31}{8} \)

Thanks for the reply Mel, though not what I wanted to hear. I hoped in 12 months it would have been fixed! 😠

A forum is only as good as its search function! 

😈 😈

Thanks for the reply Mel. I was hoping in 12 months this had been fixed. 😠

A forum is only as good as its search function!  

😈o

Yup, we await enlightenment.

To clarify the above a little:

The black triangle is a right-angled triangle lying in the x-y plane, with vertices (1,1), (4,1) and (1,2).  Its hypotenuse has the equation y = -x/3 + 7/3.  

I should really have just said that as x goes from 1 to 4, y goes from 1 to -x/3 + 7/3.

I assume this means 

A = k/3,  B = k/5 and C = k/2  where k is a constant.

A+B+C = 180°  so  k/3 + k/5 + k/2 = 180 or 31k/30 = 180 so that k = 5400/31

Hence A = 1800/31 ≈ 58.1°,  B = 1080/31 ≈ 34.8° and C = 2700/31 ≈ 87.1°

.

"Find the volume of the solid under the surface z=xy and above the triangle with verticies (1,1) (4,1) and (1,2)."

The following should help:

.

(cos11+sin11)^-2=-1

\(\frac{1}{(cos11+sin11)^2}=-1\)

This statement must be false since the square of a real number cannot be negative

Now i am playing :)

\(\frac{1}{cos^211+sin11^2+2sin11cos11}=1\\ \frac{1}{1+2sin11cos11}=1\\ \frac{1}{1+sin22}=1\\ 1=1+sin22\\ 0=sin22\;\;\;false\)

Well i am still confused.  Are you saying that z is one?  It does not say that in the question.  The way it is the actual question makes no sense.  Not to me anyway.

Why didn't you say so   LOL

\(\begin{array}{rcl}\\ax^2+bx+c&=&0\\ ax^2+bx&=&-c\\ x^2+\frac{b}{a}x&=&-\frac{c}{a}\\ x^2+\frac{b}{a}x+(\frac{b}{2a})^2&=&-\frac{c}{a}+(\frac{b}{2a})^2\\ x^2+\frac{b}{a}x+(\frac{b}{2a})^2&=&-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ (x+\frac{b}{2a})^2&=&\frac{b^2-4ac}{4a^2}\\ x+\frac{b}{2a}&=&\frac{\pm\sqrt{b^2-4ac}}{2a}\\ x&=&\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \end{array}\)

simplify square root of x^11

\(\sqrt{x^{11}}=(x^{11})^{1/2}=x^{11*1/2}=x^{11/2}=x^{5.5}\)

I don't know that you can - the search function is rubbish.   

THAT is slick....I did not know that.  That is good for all of these types of questions where someone acts like they do not know how to use a calculator yet they can spend 10 minutes entering a goofy question !

The backslash "\"  is  💦sometimes💦 used to denote integer division, e.g.,

 18\4 = 4

But often it's just someone who's dislexic and meant to type divide "/" symbol. 😯

🚳 🚴 🚵 🚷 

\(\sqrt{{x}^{11}}\)

\(\sqrt{x\times x\times x\times x\times x\times x\times x\times x\times x\times x\times x}\)

\(x\sqrt{x\times x\times x\times x\times x\times x\times x\times x\times x}\)

\({x}^{2}\sqrt{x\times x\times x\times x\times x\times x\times x}\)

\({x}^{3}\sqrt{x\times x\times x\times x\times x}\)

\({x}^{4}\sqrt{x\times x\times x}\)

\({x}^{5}\sqrt{x}\)

\({x}^{5}{x}^{\frac{1}{2}}\)

\({x}^{5+\frac{1}{2}}\)

\({x}^{\frac{10}{2}+\frac{1}{2}}\)

\({x}^{\frac{11}{2}}\)

sqrt(x^11) = sqrt(x^10* x)= x^5 sqrt(x)

So, I ended up getting it. A volume of revolution is better suited for a cone like object. Consider this a solid with the base as the triangle and the height as the z function. At point (1,1) for instance the height, z is 1. At (4,1) the height is 4. 

So what you do is take a double integral of the z function, and you bound it by the x, then the y variable (or y then x).

If you choose the outer variable to be y then the integral is from 1 to 2. X would then be from 1 to the point on the triangle where the height is y.

This is the line x=7-3y 

Thus when y = 1 x=4

It ends up being this.

\(\int_{1}^{2}\int_{1}^{7-3y}xydxdy\)

\(\int_{1}^{2}(0.5y(7-3y)^2-0.5y)dy\)

If you switch it, (which is numerically the same)

\(\int \int dydx\)

you get fractions, since y=7/3-x/3 which then has to be squared, and then a definate integral found from 1 to 4. Which I couldn't get to work, and why I ended up asking lol.

Thanks.

Nice job GJ....I just solved the equation for 'x' as requested...   Obviously x=0 will be 'undefined' with a '0' in the denominator........Which only says the equation is INVALID for x=0 (but that is not what they asked).......I did solve for y too, but did not post it,  YOU did !  

Strong work!

If you subsitute x with 0, you will find that 0 is not a solution.