MHB Good day, Exam Integrals: volume and area

[andrucabezas]

1#Find the area of the region, enclosed by:

2#Find the area of the region bounded by:

3#in the region limited by:
find the solid volume of revolution that is generated by rotating that region about the x axis

 

[HOI]

So "Do my homework for me"? No, that is not going to happen! (You don't even say "please"!)

Have you graphed these so you can see what region you are working with and what the limits of integration should be? The first problem has one boundary $x^2+ y= 4$. That is the same as $y= 4- x^2$, a parabola. its "vertex" is at (0, 4) and it crosses the x-axis at (2, 0) and (-2, 0). The next is y+ x- 2= 0 which is the same as y= 2- x. That's a straight line. When x= 0, y= 2- 0= 2 and when y= 0, 2- x= 0 so x= 2. Draw the straight line through (0, 2) and (2, 0). Do you see that the line crosses the parabola where $y= 4- x^2= 2- x$. $x^2- x- 2= (x- 2)(x+ 1)= 0$ so the line crosses the parabola at (-1, 3) and (2, 0). The other two boundaries are the vertical lines x= -2 and x= 3.

Frankly that makes no sense at all! The last two vertical lines are completely outside the region bounded by the first two so there is NO region bounded by these four. If I were forced to give a numeric answer I would give the area of the region bounded by $y= 4- x^2$ and $y= 2- x$. That region goes from x= -1 on the left to x= 2 on the right and, for each x. from y= 2- x below to $y= 4- x^2$.

Imagine dividing that region into thin vertical strips of width "dx". The length of each strip is $4- x^2- (2- x)= 2+ x- x^2$ so each strip has area $(2+ x- x^2)dx$. integrate that from x= -1 to x= 2.

 

[andrucabezas]

Lo siento, no hablo inglés con fluidez, por eso traté de escribir poco

 

[andrucabezas]

thanks for clarifying some doubts

So "Do my homework for me"? No, that is not going to happen! (You don't even say "please"!)

Have you graphed these so you can see what region you are working with and what the limits of integration should be? The first problem has one boundary $x^2+ y= 4$. That is the same as $y= 4- x^2$, a parabola. its "vertex" is at (0, 4) and it crosses the x-axis at (2, 0) and (-2, 0). The next is y+ x- 2= 0 which is the same as y= 2- x. That's a straight line. When x= 0, y= 2- 0= 2 and when y= 0, 2- x= 0 so x= 2. Draw the straight line through (0, 2) and (2, 0). Do you see that the line crosses the parabola where $y= 4- x^2= 2- x$. $x^2- x- 2= (x- 2)(x+ 1)= 0$ so the line crosses the parabola at (-1, 3) and (2, 0). The other two boundaries are the vertical lines x= -2 and x= 3.

Frankly that makes no sense at all! The last two vertical lines are completely outside the region bounded by the first two so there is NO region bounded by these four. If I were forced to give a numeric answer I would give the area of the region bounded by $y= 4- x^2$ and $y= 2- x$. That region goes from x= -1 on the left to x= 2 on the right and, for each x. from y= 2- x below to $y= 4- x^2$.

Imagine dividing that region into thin vertical strips of width "dx". The length of each strip is $4- x^2- (2- x)= 2+ x- x^2$ so each strip has area $(2+ x- x^2)dx$. integrate that from x= -1 to x= 2.

 

[andrucabezas]

I do not find logic to the statement, seeing that the x do not touch, I suppose it would have been the teacher's error, I will take the one from x = -1 to x = 2[QUOTE = "Country Boy, publicación: 122309, miembro: 11871"]
Entonces, ¿"Haz mi tarea por mí"? ¡No, eso no va a pasar! (¡Ni siquiera dices "por favor"!)

¿Ha graficado estos para que pueda ver en qué región está trabajando y cuáles deberían ser los límites de integración? El primer problema tiene un límite $ x ^ 2 + y = 4 $. Eso es lo mismo que $ y = 4- x ^ 2 $, una parábola. su "vértice" está en (0, 4) y cruza el eje x en (2, 0) y (-2, 0). El siguiente es y + x- 2 = 0, que es lo mismo que y = 2- x. Esa es una línea recta. Cuando x = 0, y = 2- 0 = 2 y cuando y = 0, 2- x = 0 entonces x = 2. Dibuja la línea recta a través de (0, 2) y (2, 0). ¿Ves que la línea cruza la parábola donde $ y = 4- x ^ 2 = 2- x $. $ x ^ 2- x- 2 = (x- 2) (x + 1) = 0 $ entonces la línea cruza la parábola en (-1, 3) y (2, 0). Los otros dos límites son las líneas verticales x = -2 y x = 3.

Francamente, eso no tiene ningún sentido. Las dos últimas líneas verticales están completamente fuera de la región delimitada por las dos primeras, por lo que NO hay ninguna región delimitada por estas cuatro. Si me viera obligado a dar una respuesta numérica, daría el área de la región limitada por $ y = 4- x ^ 2 $ y $ y = 2- x $. Esa región va de x = -1 a la izquierda a x = 2 a la derecha y, para cada x. desde y = 2- x abajo hasta $ y = 4- x ^ 2 $.

Imagine dividir esa región en delgadas franjas verticales de ancho "dx". La longitud de cada tira es $ 4- x ^ 2- (2- x) = 2+ x- x ^ 2 $ por lo que cada tira tiene un área $ (2+ x- x ^ 2) dx $. integre eso de x = -1 ax = 2.
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