Answer to Question #150520 in Calculus for Randal Rodriguez

Question

* Identify the curves (lines, parabola, etc.) generated by illustrating the graphs

5. Find the volume of the solid that lies under the surface z = xy and above the triangle with vertices (1, 1), (4,1), and (1, 2).

6. Evaluate \:\int _0^1\int _{x^2}^{\sqrt{x}}\:x^2ydydx and find the volume.

Answer 1

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

Remember that:

Using the given points, graph the triangle. Note that x lies from 1 to 4 inclusive and y lies from 1 to the line connecting the points (1,2), (4,1), By inspection, this region is a type 1. this given following equation

y - 2 = -(x-1) / 3;

y = x / 3 + 7 / 3;

Set D = (x, y) | 1 $\leq$ x $\leq$ 4, 1 $\leq$ y $\leq$ -x/3 + 7/3;

V = $\:\int _1^4\int _{1}^{\frac{-x}{3}+\frac{7}{3}}\:xydydx$ = $\:\int _1^4\lbrack \frac{xy^2}{2} \rbrack ^ {\frac{-x}{3}+\frac{7}{3}}dx$ =

$\frac{1}{2}\:\int _1^4\lparen x \lparen \frac{-x}{3} + \frac{7}{3} \rparen - x \rparen dx$ =

= $\frac{1}{2}\:\int _1^4\lparen x^3 - 14x^2+40x \rparen dx$ = $\frac{1}{18} \lbrack \frac{x^4}{4} - \frac{14x^3}{3} + 20x^2 \rbrack _1^4 =$

$\frac{1}{18} \lparen\frac{1}{4}\lparen255\rparen-\frac{14}{3}\lparen63\rparen+20\lparen15\rparen\rparen =\frac{31}{8}$

1. Evaluate $\:\int _0^1\int _{x^2}^{\sqrt{x}}\:x^2ydydx$ and find the value

V = $\:\int _0^1\:0.5x^2y^2\text{\textbar}_{x^2}^{\sqrt{x}}dx$ = $\:\int _0^1\:0.5x^2(\sqrt{x})^2-0.5x^2(x^2)^2dx$ = $\:\int _0^1\:0.5(x^3-x^6)dx$ =

$x^4/8 - x^7/14\text{\textbar}_0^{1}$ = 1 / 8 - 1 / 14 = (7 - 4) / 56 = 3 / 56;